problem solving examples in displacement

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Displacement and distance: tutorials with examples.

Examples with explanations on displacement and distance of objects moving along straight lines. More problems and their solution can be found in this site.

Distance and Displacement Definitions

The distance is a scalar quantity (magnitude) that describes the length of the total path covered by a moving object. The displacement is a vector quantity (magnitude and direction) that describes the difference between the final and initial positions of a moving object. It is the shortest distance moved in a certain direction. Both distance and displacement are measured in unit of lengths. (centimeters, meters, kilometers,...)

Examples with Detailed Solutions

More References and links

• Velocity and Speed: Tutorials with Examples
• Velocity and Speed: Problems with Solutions
• Acceleration: Tutorials with Examples
• Uniform Acceleration Motion: Problems with Solutions
• Uniform Acceleration Motion: Equations with Explanations

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Distance and Displacement

• A mountain climbing expedition establishes a base camp and two intermediate camps, A and B. Camp A is 11,200 m east of and 3,200 m above base camp. Camp B is 8400 m east of and 1700 m higher than Camp A. Determine the displacement between base camp and Camp B.
• A laser beam is aimed 15.95° above the horizontal at a mirror 11,648 m away. It glances off the mirror and continues for an additional 8570. m at 11.44° above the horizon until it hits its target. What is the resultant displacement of the beam to the target?
• one complete orbit around the Sun
• one-half orbit around the Sun
• one-fourth orbit around the Sun
• What relationship between distance and displacement does this illustrate?
• How far do you walk each day?
• How far is it from Toronto to Mexico City by airplane?
• How far is it from Toronto to Mexico City by car?
• How far is the Earth from the moon?
• How long is a standard sheet of plywood?
• How long is the coastline of Britain?
• How long is the Nile?
• How long is the track on a compact disc?
• How long did it take you to get here?
• How many frequent flyer miles do you have?
• What is the radius of the Earth?
• An ambitious hiker walks 25 km west and then 35 km south in a day. Find the magnitude and direction of the hiker's resultant displacement (relative to due west).
• What distance did I travel?
• What's my resultant displacement (magnitude and direction relative to due east)?
• How far would Lois Lane have to walk to reach Superman?
• How far would Superman have to fly to reach Lois Lane?
• In what direction should Superman fly to reach Lois Lane as quickly as possible?

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1.2: Displacement

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Learning Objectives

• Define position, displacement, distance, and distance traveled.
• Explain the relationship between position and displacement.
• Distinguish between displacement and distance traveled.
• Calculate displacement and distance given initial position, final position, and the path between the two.

In order to describe the motion of an object, you must first be able to describe its  position —where it is at any particular time. More precisely, you need to specify its position relative to a convenient reference frame. Earth is often used as a reference frame, and we often describe the position of an object as it relates to stationary objects in that reference frame. For example, a rocket launch would be described in terms of the position of the rocket with respect to the Earth as a whole, while a professor’s position could be described in terms of where she is in relation to the nearby white board. (See  Figure  \(\PageIndex{2}\).) In other cases, we use reference frames that are not stationary but are in motion relative to the Earth. To describe the position of a person in an airplane, for example, we use the airplane, not the Earth, as the reference frame. (See  Figure  \(\PageIndex{3}\).)

Displacement

If an object moves relative to a reference frame (for example, if a professor moves to the right relative to a white board or a passenger moves toward the rear of an airplane), then the object’s position changes. This change in position is known as  displacement . The word “displacement” implies that an object has moved, or has been displaced.

Definition: DISPLACEMENT

Displacement is the  change in position  of an object:

\[\Delta x=x_{\mathrm{f}}-x_{0}, \nonumber\]

where \(\Delta x\) is displacement, \(x_{\mathrm{f}}\) is the final position, and \(x_{0}\) is the initial position.

In this text the upper case Greek letter \(\Delta\) always means “change in” whatever quantity follows it; thus, \(\Delta x\) means  change in position . Always solve for displacement by subtracting initial position \(x_{0}\) from final position \(x_{f}\).

Note that the SI unit for displacement is the meter (m), but sometimes kilometers, miles, feet, and other units of length are used. Keep in mind that when units other than the meter are used in a problem, you may need to convert them into meters to complete the calculation.

Note that displacement has a direction as well as a magnitude. The professor’s displacement is 2.0 m to the right, and the airline passenger’s displacement is 4.0 m toward the rear. In one-dimensional motion, direction can be specified with a plus or minus sign. When you begin a problem, you should select which direction is positive (usually that will be to the right or up, but you are free to select positive as being any direction). The professor’s initial position is \(x_{0}=1.5 \mathrm{~m}\) and her final position is \(x_{\mathrm{f}}=3.5 \mathrm{~m}\). Thus her displacement is

\[\Delta x=x_{\mathrm{f}}-x_{0}=3.5 \mathrm{~m}-1.5 \mathrm{~m}=+2.0 \mathrm{~m} . \nonumber\]

In this coordinate system, motion to the right is positive, whereas motion to the left is negative. Similarly, the airplane passenger’s initial position is \(x_{0}=6.0 \mathrm{~m}\) and his final position is \(x_{\mathrm{f}}=2.0 \mathrm{~m}\), so his displacement is

\[\Delta x=x_{\mathrm{f}}-x_{0}=2.0 \mathrm{~m}-6.0 \mathrm{~m}=-4.0 \mathrm{~m} . \nonumber\]

His displacement is negative because his motion is toward the rear of the plane, or in the negative \(x\) direction in our coordinate system.

Although displacement is described in terms of direction, distance is not.  Distance  is defined to be  the magnitude or size of displacement between two positions . Note that the distance between two positions is not the same as the distance traveled between them.  Distance traveled  is  the total length of the path traveled between two positions . Distance has no direction and, thus, no sign. For example, the distance the professor walks is 2.0 m. The distance the airplane passenger walks is 4.0 m.

MISCONCEPTION ALERT: DISTANCE TRAVELED VS. MAGNITUDE OF DISPLACEMENT

It is important to note that the  distance traveled , however, can be greater than the magnitude of the displacement (by magnitude, we mean just the size of the displacement without regard to its direction; that is, just a number with a unit). For example, the professor could pace back and forth many times, perhaps walking a distance of 150 m during a lecture, yet still end up only 2.0 m to the right of her starting point. In this case her displacement would be +2.0 m, the magnitude of her displacement would be 2.0 m, but the distance she traveled would be 150 m. In kinematics we nearly always deal with displacement and magnitude of displacement, and almost never with distance traveled. One way to think about this is to assume you marked the start of the motion and the end of the motion. The displacement is simply the difference in the position of the two marks and is independent of the path taken in traveling between the two marks. The distance traveled, however, is the total length of the path taken between the two marks.

Exercise \(\PageIndex{1}\)

A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is her displacement? (b) What distance does she ride? (c) What is the magnitude of her displacement?

Figure \(\PageIndex{4}\)

Section Summary

• Kinematics is the study of motion without considering its causes. In this chapter, it is limited to motion along a straight line, called one-dimensional motion.
• Displacement is the change in position of an object.
• When you start a problem, assign which direction will be positive.
• Distance is the magnitude of displacement between two positions.
• Distance traveled is the total length of the path traveled between two positions.

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Displacement and Distance with Examples

In this article, examples of displacement and distance in physics are presented along with their definitions which is helpful for high school physics.  

Definition of displacement and distance

Displacement is a vector quantity describes a change in position of an object or how far an object is displaced from its initial position and is given by below formula  \[ \Delta \vec x = x_f - x_i \] Where the starting and ending positions are denoted by $x_i$ and $x_f$, respectively.  Distance is a scalar quantity indicating the total path traveled by a moving object.

Since in both cases, the interval between two points is measured- one as the shortest line and the other as the total traveled path- so the SI unit of displacement and distance is the meter.  The illustration below shows the difference between them clearly.

5 Examples of displacement and distance

Now with this brief definition, we can go further and explain those concepts more concisely with numerous examples.

Example (1): A boy is playing around a rectangular path. He starts his game from one corner ($i$) and ends at the same one. 

Solution: In this path, since his initial and final points are the same, by definition, his displacement is zero .

On the other hand, the total distance traveled by him is the perimeter of that rectangular shape. 

In this way, distance traveled (in physics) is consistent with the concept of displacement in everyday language. 

Example: A car travels a circle of the circumference of $100\,\rm m$. In each of the following, find the quantity wanted: (a) If the car moves once around the track, what is the distance traveled by it? What if it moves twice? (b) If the car moves once around the track, what is its displacement?

Solution : Another example of displacement that clarifies its difference with the distance traveled.

Distance is defined as the whole path covered by a moving object, but displacement is the difference between the initial and final positions of the moving body. 

(a) The whole path taken by the car around the circle is its circumference. Thus, when the car travels once around the circle, its distance traveled is $100\,\rm m$.

Once the car moves twice around the circle, its distance traveled is $200\,\rm m$. 

(b) When the car travels a complete circle, its initial and final positions coincide with each other. Thus, according to the definition of displacement, $\Delta x=x_f-x_i$, its displacement becomes zero.  

This example explicitly shows us that the distance traveled cannot never get zero.

Example (2): A biker ride into a horizontal loop of a radius of 10 meters and covers three fourth of it as follows. What are the displacement and distance traveled by him?  

Solution: As mentioned earlier, displacement is a quantity that depends only on the position of initial and final points.

Thus, the straightest line between those points, which gives the magnitude of the displacement vector, is computed by Pythagorean theorem as \begin{align*} D^2 &= r^2 + r^2 \\ &= 2 r^2 \\ \Rightarrow D &= r\sqrt{2} \end{align*} where the magnitude of displacement denoted by $D$. Putting the values gives, $D=10\sqrt{2}\,\rm m$. Distance is simply the circumference of three fourth of the circle, so \begin{align*} \text{distance}&= \frac{3}{4} \, {\text{perimeter}} \\\\ &=\frac{3}{4} (2\pi\,r) \\\\ &= \frac{3 \times 2\times \pi \times 10}{4} \\\\  &=15\,\pi \,\rm m \end{align*}

Example (3): A person starts at position 0 and walks 4 meters to the right, returns, and walks 6 meters to the left.  (a) What is displacement? (b) What is the distance traveled by him?

Solution : denote the initial position as $x_i=0$ and the final position as $x_f$. As you can see, from the starting position to the turning point, the person walks 4 meters to the right. From that point ($B$) count 6 steps to the left to reach the final point whose position is on the negative side of the axis i.e. $x_f=-2$. 

(a) Displacement in physics is a vector and is defined to be the final position minus the initial position. So, \[\Delta x=x_f-x_i=-2-0=-2\,{\rm m}\] Hence, the person displaced 2 meters to the left (green arrow). Here, the minus sign indicates the direction of the displacement. 

(b) For the first part of the trip, the person walks 4 meters, and next, 6 meters more additional to the left. Thus, the total distance traveled is calculated to be 10 meters.

Example (4): If you walk exactly 4 times around a quarter-meter path, what is your displacement?

Solution : let the initial position be $x_i=0$. By gluing 4 of these quarter tracks, you reach your initial position. So, the initial and final positions are the same and consequently, the displacement becomes zero.

Example (5): Consider a car to be a point particle and move in one dimension. To specify the location of the particle in one dimension we need only one axis that we call $x$ and lies along the straight-line path.

Solution: First, we must define an important quantity that the other kinematics quantities are made from it, displacement.

To describe the motion of the car, we must know its position and how that position changes with time.

The change in the car's position from initial position $x_i$ to final position $x_f$ is called displacement, $\Delta \vec x= x_f-x_i$ (in physics we use the Greek letter $\Delta$ to indicate the change in a quantity).

This quantity is a vector  point from $A$ to $B$ and in $1$-D denoted by $\Delta \vec x=x_B-x_A$.

In the figure below, the car moves from point $A$ at $x=2\, {\rm m}$ and after reaching to $x=9\,{\rm m}$ returns and stops at position $x=6\,{\rm m}$ at point $B$.

Therefore, the car's displacement is $\Delta x=6-2=+4\,{\rm m}$.

Another quantity which sometimes confused with displacement is the distance traveled  (or simply distance ) which is defined as the overall distance covered by the particle.

In the above example, the distance from the initial position is computed as follows:

First, calculate the distance to the return point $d_1=x_C-x_A=9-2=7\,{\rm m}$ then from that point ($x_C$) to the final point $x_B$ i.e. $d_2=x_B-x_C=6-9=-3$.

But we should pick the absolute value of it since the distance is a scalar  quantity and for them, a negative value is non-sense.

Therefore, the total distance covered by our car is $d_{tot}=d_1+|d_2|=7+|-3|=7+3=10\,{\rm m}$ .

In case of several turning points along the straight path or once the motion's path is on a plane or even three-dimensional cases, one should divide the overall path (one, two, or three dimensions) into straight lines (without any turning point), calculate the difference of those initial and final points and then add their absolute values  of each path to reach to the distance traveled by that particle on that specific path (see examples below). 

Displacement in two and three dimensions

In more than one dimension, the computations are a bit involved and we need to be armed with additional concepts.

In this section, we can learn how by using vectors one can describe the position of an object, and by manipulating them to characterize the displacement and other related kinematical quantities (like velocity and acceleration ). 

In a coordinate system, the position of an object is described by a so-called position vector  that extends from reference origin $O$ to the location of the object $P$ and is denoted by $\vec{r}=\overrightarrow{OP}$.

These vectors, in the Cartesian coordinate system (or other related coordinates), can be expressed as a linear combination of unit vectors, $\hat{i},\hat{j},\hat{k}$, (the ones with unit length) as  \[ \textbf{r}=\sum_1^{n} r_x \hat{i}+r_y \hat{j} +r_z \hat{k} \] where $n$ denotes the dimension of the problem, i.e., in two and three dimensions $n=2,3$, respectively. $r_x , r_y$ and $r_z$ are called the components of the vector $\vec{r}$.

Now the only thing that remains is adding or subtracting these vectors, known as vector algebra, to provide the kinematical quantities.

To do this, simply add or subtract the terms (components) along a specific axis with each other (as below).

Consider adding two vectors $\vec{a}$ and $\vec{b}$ in two dimensions, \begin{array}{cc} \textbf{a}+\textbf{b}&=&\left(a_x \hat{i}+a_y \hat{j} \right)+\left(b_x \hat{i}+b_y \hat{j}\right)\\ &=&\left(a_x+b_x\right)\hat{i}+\left(a_y+b_y\right)\hat{j}\\ &=&c_x\, \hat{i}+c_y\, \hat{j} \end{array}In the last line, the components of the final vector (or resultant vector) are denoted by $c_x$ and $c_y$.

As we learned from vector practice problems , the magnitude and direction of the obtained vector are represented by the following relations \begin{array}{cc} |\textbf{c}| &=& \sqrt{\left(a_x+b_x\right)^2 +\left(a_y+b_y\right)^2 } \qquad \text{magnitude}\\ \theta &=& \tan^{-1} \left(\frac{a_y+b_y}{a_x+b_x}\right) \qquad \text{direction} \end{array} where $\theta$ is the angle with respect to the $x$ axis.

We have two types of problems on the topic of displacement.

In the first case, the initial and final coordinates (position) of an object are given.

Write position vectors for every point. The vector which extends from the tail of the initial point to the tail of the final point is a displacement vector and computed as the difference of those vectors i.e. $\vec{c}=\vec{b}-\vec{a}$.

In the second case, the overall path of an object, between the initial and final points, is given as consecutive vectors as in the figure below.

Here, one should decompose each vector with respect to its origin, then add components along $x$ and $y$ axes separately.

The displacement vector is the one that points from the tip of the first vector to the tail of the last vector and its magnitude is the vector addition of those vectors i.e. $\vec{d}=\vec{a}+\vec{b}+\vec{c}$. 

More examples of displacement:

Example 1: A moving object is displaced from A(2,-1) to B(-5,3) in a two-dimensional plane. What is the displacement vector of this object?

Solution : First, this is the first case mentioned above. So construct the position vectors of point $A$ and $B$ as below \begin{gather*} \overrightarrow{OA}=2\,\hat{i}+(-1)\,\hat{j}\\ \overrightarrow{OB}=-5\,\hat{i}+3\,\hat{j} \end{gather*} Now, by definition, the difference of initial and final points or simply position vectors gets the displacement vector $\vec{d}$ as \begin{align*}\vec{d} &=\overrightarrow{OB}-\overrightarrow{OA}\\\\ &= \left(-5\,\hat{i}+3\,\hat{j}\right)-\left(2\,\hat{i}+(-1)\,\hat{j}\right)\\\\ &= -7\,\hat{i}+4\,\hat{j} \end{align*} its magnitude and direction is also obtained as follows \begin{align*} |\vec{d}|&=\sqrt{(-7)^2 +4^2} \\\\ &=\sqrt{49+16} \\\\ &\approx 8.06\,{\rm m} \end{align*} and \[\theta = \tan^{-1} \left(\frac{d_y}{d_x}\right)=\tan^{-1}\left(\frac{4}{-7}\right)\] The angle $\theta$ may be $-29.74^\circ$ or $150.25^\circ$ but since $d_x$ is negative and $d_y$ is positive so the resultant vector lies on the second quarter of coordinate system. Therefore, the desired angle with $x$-axis is $150.25^\circ$.

Example (2): An airplane flies $276.9\,{\rm km}$ $\left[{\rm W}\, 76.70^\circ\, {\rm S}\right]$ from Edmonton to Calgary and then continues $675.1\,{\rm km}$ $\left[{\rm W}\, 11.45^\circ\,{\rm S}\right]$ from Calgary to Vancouver. Using components, calculate the plane's total displacement. (Nelson 12, p. 27). 

Solution : In these problems, there is a new thing that appears in many textbooks which is the compact form of direction as stated in brackets. $\left[{\rm W}\, 76.70^\circ\, {\rm S}\right]$ can be read as "point west, and then turn $76.70^\circ$ toward the south".

Example (3): A moving particle moves over the surface of a solid cube in such a way that passes through $A$ to $B$. What is the magnitude of the displacement vector in this change of location of the particle?

Solution : In three-dimensional cases like $2-D$ ones, we should only know the location(coordinates) of the object, then use the following relations to obtain the displacement of a moving particle.Points $A$ and $B$ lies on the $x-z$ plane and $y$ axis, respectively so their coordinates are $(10,0,10)$ and $(0,10,0)$ which parenthesis denote the $(x,y,z)$. This question is of type one, so \begin{align*} \overrightarrow{OA}&= 10\,\hat{i}+0\,\hat{j}+10\,\hat{k} \\ \overrightarrow{OB} &= 0\,\hat{i}+10\,\hat{j}+0\,\hat{k} \end{align*} and displacement vector is \begin{align*} \vec{d} &=\overrightarrow{OB}-\overrightarrow{OA} \\ &= -10\,\hat{i}+10\,\hat{j}-10\,\hat{k} \end{align*} Therefore, the wanted vector in terms of its components was computed as above. Also, its magnitude is the square root of the sum of the squares of each component, i.e., \[|\vec{d}|=\sqrt{(-10)^2 +(-10)^2 + (10)^2}=10\sqrt{3}\rm m\]

Example (4): A car moves around a circle of radius of $20\,{\rm m}$ and returns to its starting point. What is the distance and displacement of the car? ($\pi = 3$)

Solution : As mentioned above, displacement depends on the initial and final points of the motion. Because the car returns to its initial position so no displacement is made by the car. But the amount of distance traveled is simply the perimeter of the circle (since this scalar quantity depends on the form of the path). So $d=2 \pi r=2 \times 3 \times 20 =120\,{\rm m}$, where $r$ is radius of the circle.

Example (5): A moving object is covering a square path -with one end left open- as shown in the figure below. What is the desired displacement and distance traveled between the specified points? point $p$ lies in the middle of $BC$.

Solution:  Displacement is the shortest and straightest line between the initial and final points.

So using the Pythagorean theorem, we get \begin{align*} D^2 &= (iB)^2 +(Bf)^2 \\\\ &= 5^2+(2.5)^2 \\\\ \Rightarrow D &= \sqrt{25 + 6.25} \\\\ &\approx 5.6\quad (\rm m) \end{align*} The direction of the displacement vector is also obtained as \begin{align*} \tan \theta &= \frac{Bp}{iB} \\\\ &= \frac{2.5}{5} \\\\ \Rightarrow \theta &= \arctan \frac{2.5}{5} \\\\ &= 26.5^\circ\, \left[\text{South east}\right] \end{align*} Distance is simply the perimeter of the  path traveled, so \begin{align*} \text{distance} &=5 + 2.5 \\ &=7.5\quad \rm m \end{align*}

Example (6):  you walk once around an oval track for 6.0 minutes at an average speed of 1.7 m/s.

(a) What is the distance traveled?

(b) What is the displacement? 

Solution :(a) here, the distance covered by the person is the perimeter of the ellipse. Since the time elapsed and average speed are given, we can use the definition of average speed and solve for the distance traveled to find it as below \begin{align*} \text{average speed}&=\frac{\text{distance}}{\text{time interval}}\\1.7&=\frac{d}{6\times 60\,{\rm s}}\\ &\\ \Rightarrow d&=\big(1.7\,{\rm \frac ms}\big)(360\,{\rm s})\\&=612\quad {\rm m} \end{align*}

(b) Displacement is the difference between initial and final positions. In this problem, since the person had returned to his original position so his displacement is zero . 

Example (7): The velocity versus time graph of a car is shown below. Find the displacement in 4 seconds.

Solution : The area under a velocity-time graph in a given time interval indicates the displacement in that interval. 

In this case, the bounded area is a triangle whose base is 5 units and height is 2 units. So, its area is \begin{align*} \text{area=displacement}&=\frac 12 \times base\times height\\\\&=\frac 12 \times 5\times 2\\\\&=5\quad {\rm m}\end{align*} Thus, this car displaces 5 meters in 5 s.

As you can see in the examples of displacement above, displacement is a vector that depends only on the initial and final positions of the object, and not on the details of the motion and path.

These vector quantities require both a length and a direction to be identified.

It is also possible to find the displacement vector using a position vs. time graph .

On the contrary, distance is a quantity that is characterized only by a simple value, which is called scalar,  and is path dependence.

In general, the distance traveled and the magnitude of the displacement vector between the two points is not the same.

If the moving object changes its direction in the course of travel, then the total distance traveled is greater than the magnitude of the displacement between those points.

The SI units of both quantities are meters.

Distance and displacement are frequently used in solving velocity and acceleration problems .

Author :  Ali Nemati

Date Published: 26 July 2017

Last Update : 8-28-2021

© 2015 All rights reserved. by Physexams.com

Distance & Displacement – Definitions, Differences and Worksheets

• 1.1 Distance and Displacement – Definitions
• 1.2 Features of displacement
• 1.3 Difference between distance and displacement in tabular form
• 1.4 Distance and Displacement problems

By going through this article, you will get a thorough knowledge about the distance as well as the displacement, and how to find it from the graphs, with the help of practice problems.

Distance and Displacement – Definitions

The total path covered by a particle in a given interval of time is called the distance travelled by that particle.

Distance is a scalar quantity. For a moving body, the distance covered is always positive. The SI unit for distance is metre and its dimensional formula is [M 0 L 1 T 0 ].

Consider a particle moving from P to R via Q as shown in figure 1. Then, the distance covered = PQ + QR

In the case of a body circulating around a circle of radius r, when it completes a semi-circle AB, the distance covered is πr, while displacement is 2r. When it completes one revolution, the distance covered is 2πr and the displacement is zero.

• Consider an object projected vertically from the top of a building of height h. The object moves a distance l upwards and returns to the ground. Now the distance travelled is (2l + h) while the displacement is –h.

Features of displacement

• Displacement is a vector quantity. It can be a positive,a negative or a zero value.
• The magnitude of displacement of a particle between two points gives the shortest distance between the two points.
• The SI unit for displacement is metre and it has the dimension of length [M 0 L 1 T 0 ]
• The magnitude of the displacement will be less than or equal to the actual distance travelled by the object in the given interval of time. i.e., |Displacement| ≤ Distance
• If an object, after travelling a certain distance returns to the starting point, then its displacement is zero.
• The displacement of the object between two points has a unique value.
• The displacement of an object is unaltered due to a shift in the origin of the position axis.
• The numerical ratio of displacement to distance is equal to or less than one.

Difference between distance and displacement in tabular form

Distance and displacement problems.

Problem 1: A boat sailing through a river moved eastward for 5 km, then cross the river by moving 3 km southward. On reaching the other side it moved westward through 1 km and reached the jetty. Find the distance covered and displacement of the boat.

Solution: Figure 3 From the fig,  AB = 5 km                        ED = BC = 3 km                        CD = 1 km                     ∴ AE = 5 km – 1 km = 4 km Distance covered = AB + BC + CD                              = 5km + 3km + 1 km                              = 9 km So the distance covered by the boat is 9 km and the displacement is 5 km

Problem 2: A car moving along in a straight highway from point P to point Q to point R and to point S, then back to point Q and finally to the point R as shown in the figure below. a) Find the distance travelled by car. b) Find the displacement of the car.

Ans: Figure 4 Given the distances, PQ = 3 km, QR = 5 km and RS = 7 km Also, SQ = 7 + 5 =12 km, PR = 3 + 5 = 8 km a) Distance travelled by the car = PQ + QR + RS + SQ + QR                                                     = 3 + 5 + 7 + 12 + 5                                                     = 32 km b) Displacement of the car = the shortest distance between the final point R and the initial point P                                             = PR                                             = 8 km

Problem 3: A person walks along the path of a rectangle from point P to point R as shown in the below figure. a) Find the distance travelled by the person. b) Find out the magnitude of the displacement of the person.

Ans: Given the distances, PQ = 5 km,                                  QR = 2 km a) The distance travelled by the person = PQ + QR                                                                = 5 + 2                                                                = 7 km b) The magnitude of displacement is equal to the shortest distance between the final point R and the initial point P, which is equal to the diagonal PR and can be calculated using Pythagora’s theorem. i.e.,  magnitude of displacement = PR By Pythagora’s theorem,    PR 2 = PQ 2 + QR 2

Problem 4: A motorcycle rides from point P to Q to R to S and finally to P in a circular path as shown in the below figure. Find a) the distance travelled by motorcycle. b) the displacement.

Answer: Given radius, r = 5 km a) Here, the distance travelled by motor cycle = circumference of the circle (∵ the motor cycle moves one complete rotation)                                                                          = 2πr                                                                          = 2 × 3.14 × 5                                                                          = 31.4 km b) Here, the initial point is P and the final point is also P. Therefore, there will be no change in position and hence the displacement is equal to zero.

Problem 5: A vehicle moves from point P to Q to R to S in a circular path as shown in the below figure. a) Find the distance travelled by the vehicle. b) Find out the magnitude of the displacement of the vehicle.

Ans:   Given, radius, r = 8 km a) Here, the vehicle moves only 3/4 th of one rotation. ∴, the distance travelled by the vehicle = 3/4 th of the circumference of the circle.                                                               =  (3/4)2πr                                                               = (3/4) × 2 ×  3.14 × 8                                                               = 37.68 km b) Here the initial point is P and the final point is S. ∴, The magnitude of displacement is equal to the shortest distance between the final point S and the initial point P, which is equal to the distance PS and can be calculated using Pythagoras theorem to the triangle POS as shown in the below figure.

Ans: From the distance time graph provided in the question, we can find out the distance and displacement for different time intervals. From the fig, a) The distance travelled by the object in first 2 seconds = 60 m b) The distance travelled by the object in first 4 seconds = 90 m c) The distance travelled by the object in first 6 seconds = 90 m d) The distance travelled by the object in first 8 seconds = 150 m e) Distance travelled by the object in 9 seconds = 150 + (150  – 120 )                                                                              = 150 m + 30 m                                                                             = 180m. That is, from t = 0 to t = 8 seconds, the object has moved a distance of 150 m from the origin and from t = 8 to t = 9 sec, the object has travelled back a distance of 30 meters. So the total distance will be 150 + 30 = 180 m. f) The distance travelled by the object in 14 seconds = Total distance travelled by the object.                                                                                     = 150 + 30 + 120                                                                                     = 300 m. That is, from t = 0 to t = 8 seconds, the object has moved a distance of 120 m from the origin and from t = 8 to  t = 14 sec, the object has come back to its initial position and travelled a distance of 150 meters again. So total distance travelled = 150 m + 150 m = 300 metre. g) Since the object has come back to its initial position, the total displacement is zero.

Answer: a) Consider the object P Object P had an initial position of 1 metre and a final position of 4 metres. ∴ displacement of object P, Δx p = final position – initial position                                                    = 4 – 1                                                    = +3 metres. Distance travelled by object P    = total path covered covered by P                                                    = 3 metres b) Consider the object Q Object Q had an initial position of 11 metres and a final position of 7 metres. ∴ displacement of object Q, Δx q = final position – initial position                                                     = 7 – 11                                                     = -4 metres. Distance travelled by object Q    = total path covered by Q                                                     = 4 metres c) Consider the object R Object R had an initial position of 0 metres and a final position of 6 metres. ∴ displacement of object R, Δx r = final position – initial position                                                    = 6 – 0                                                    = +6 metres. Distance travelled by object R    = total path covered by R                                                     = 6 + 3 + 3                                                     = 12 metres d) Consider the object S Object S had an initial position of 8 metre and final position of 7 metres. ∴ displacement of object S, Δx s = final position – initial position                                                     = 7 – 8                                                     = –1 metre. Distance travelled by object S     = total path covered by S                                                      = 3 + 4                                                      = 7 metres e) Consider the object T Object T had an initial position of 7 metres and final position of 8 metres. ∴ displacement of object T, Δx t  = final position – initial position                                                    = 8 – 7                                                    = 1 metre. Distance travelled by object T    = total path covered by T                                                    = 4 + 3                                                    = 7 metres.

Answer: From the position-time graph given, we can calculate the distance and displacement for different time intervals. a) Between 1 hour an 3 hour, the car had an initial position of 40 km and a final position of 40 km. ∴ Displacement of the car between 1 hour and 3 hours , Δx = final position – initial position = 40 – 40 = 0 km . Distance travelled by car between 1 hour and 3 hours = total path covered between 1 hour and 3 hours = 0 km ( Since that portion of the graph is a straight line parallel to the x-axis). b) Similarly, the car had an initial position of 40 km and a final position of 160 km between 3 hours and 5 hours. ∴ Displacement of the car between 3 hour and 5 hours , Δx = final position – initial position = 160 – 40 = 120 km . Distance covered by the car between 3 hour and 5 hours = total path covered between 3 hours and 5 hours = 120 km . c) Similarly, the car starts at an initial position of 160 km and ends at a final position of 0 km during the period 5 hours and 9 hours. ∴ Displacement of the car between 5 hours and 9 hours , Δx = final position – initial position = 0 – 160 = -160 km ( displacement is negative means the car moves in the opposite or negative direction). Distance covered by the car between 5 hours and 9 hours = total path covered between 5 hours and 9 hours = 160 km (distance is always positive) d) Similarly, between the time 3 hours and 9 hours,  the initial and final position of the car is 40 km and 0 km respectively. ∴ Displacement of the car between 3 hours and 9 hours, Δx = final position – initial position = 0 – 40 = -40 km ( displacement is negative means the car moves in the opposite or negative direction). From the graph, it is clear that the car travels in two segments between 3 hours and 9 hours. i.e., the car starts at 40 km at 3 hours and moves to 160 km at 5 hours, travelling a distance of 120 km. Also, the car starts at 120 km at 5 hours and moves to 0 km at 9 hours, travelling a distance of 160 km. ∴ Distance covered by the car between 3 hours and 9 hours = total path covered between 5 hours and 9 hours = 120 km + 160 km = 280 km. e) While considering the total motion, the car starts and ends at the same position of 0 km. That is, its initial position is 0 km at 0 hour and final position is also 0 km at 9 hours. ∴ Total displacement of the car , Δx = Displacement of the car between 0 hours and 9 hours= final position – initial position = 0 – 0 = 0 km . Also, while considering the total motion, it is clear that the car travels in four segments between 0 hours and 9 hours. i.e., the car starts at 0 km at 0 hours and moves to 40 km at 1 hour, travelling a distance of 40 km. Also, the car is stationary between the time 1 hour and 3 hours, thus the distance  covered in this segment is 0 km. Again, the car starts at 40 km at 3 hours and moves to 160 km at 5 hours, travelling a distance of 120 km. Finally, the car starts at 120 km at 5 hours and moves to 0 km at 9 hours, travelling a distance of 160 km. ∴ Total distance covered by the car = total path covered between 0 hours and 9 hours = 40 + 0 + 120 + 160 = 320 km

Answer: The distance and displacement for different time intervals can be found out from the position vs time graph given. a) The bicycle had an initial position of 60 metres and a final position of –30 metres between 0s and 30 seconds. ∴ Displacement of the bicycle between 0 second and 30 seconds , Δx = final position – initial position = -30 – 60 = -90 m. Also, from the p-t graph given, it is clear that the bicycle travels in two segments between 0 second and 30 seconds. i.e., the bicycle starts at 60 m at 0 second and moves to – 30 metres at 15 seconds, travelling a distance of 90 m. Also, the bicycle is stationary between the time 15 seconds and 30 seconds, thus the distance travelled in this segment is 0 m. ∴ Distance covered by the bicycle between 0 second and 30 seconds = total path covered between 0 seconds and 30 seconds = 90 + 0 = 90 m. b) Similarly, between the time 0 seconds and 40 seconds, the bicycle had an initial and final position of 60 metres and 30 metres respectively. ∴ Displacement of the bicycle between 0 second and 40 seconds , Δx = final position – initial position = 30 – 60 = -30 m. Also, from the graph, it is clear that the bicycle travels in three segments between 0 second and 40 seconds. i.e., the bicycle starts at 60 m at 0 second and moves to – 30 metres at 15 seconds, travelling a distance of 90 m. Also, the bicycle is stationary between the time 15 seconds and 30 seconds, thus the distance travelled in this segment is 0 m. Again, the bicycle starts at –30 metres at 30 seconds and moves to 30 metres at 40 seconds, covering a distance of 60 metres. ∴ Distance covered by the bicycle between 0 second and 40 seconds = total path covered between 0 seconds and 40 seconds = 90 + 0 + 60 = 150 m. c) Similarly, the bicycle starts at an initial position of -30 metres and ends at a final position of 0 metres during the period 30 seconds and 50 seconds. ∴ Displacement of the bicycle between 30 second and 50 seconds , Δx = final position – initial position = 0 – -30 = +30 m. Also, from the graph, it is clear that the bicycle travels in two segments between 30 second and 50 seconds. i.e., the bicycle starts at -30 m at 30 seconds and moves to 30 metres position at 40 seconds, travelling a distance of 60 m. Also, the bicycle starts at 30 metres at 40 seconds and moves to 0 metres position at 50 seconds, covering a distance of 30 metres. ∴ Distance covered by the bicycle between 0 second and 50 seconds = total path covered between 0 seconds and 50 seconds = 60 + 30 = 90 m. d) Now while considering the total motion, the bicycle starts at 60 metres and ends at 0 metres. ∴ Total displacement of the car , Δx = Displacement of the car between 0 second and 50 seconds = final position – initial position = 0 – 60 = -60 m. Also, from the graph, it is clear that the bicycle travels in four segments between 0 second and 50 seconds. i.e., the bicycle starts at 60 m at 0 second and moves to – 30 metres at 15 seconds, travelling a distance of 90 m. Also, the bicycle is stationary between the time 15 seconds and 30 seconds, thus the distance travelled in this segment is 0 m. Again, the bicycle starts at –30 metres at 30 seconds and moves to 30 metres at 40 seconds, covering a distance of 60 metres. Finally, the bicycle starts at 30 metres at 40 seconds and moves to a 0 metre position at 50 seconds, covering a distance of 30 metres. ∴ Total distance covered by the bicycle = total path covered between 0 second and 50 seconds = 90 + 0 + 60 + 30 = 180 m.

Problem 10: The position-time graph for an elevator travels up and down is given below. Find the distance and displacement of the elevator between 6 seconds and 21 seconds.

Answer: The elevator had an initial position of -15 metres and a final position of 20 metres between 6 s and 21 seconds. ∴ Displacement of the elevator between 6 seconds and 21 seconds , Δx = final position – initial position = 20 – -15 = 35 m. Also, from the graph, it is clear that the elevator moves in two segments between 6 second and 21 seconds. i.e., the elevator starts at -15 m at 6 second and moves to a 0-metre position at 9 seconds, covering a distance of 15 m. Again, the elevator starts at 0 m at 9 seconds and moves to 20 metres position at 21 seconds, covering a distance of 20 m. ∴ Distance covered by the elevator between 6 seconds and 21 seconds = total path covered between 6 seconds and 21 seconds = 15 + 20 = 35 m.

I hope the information in this article helps you to get a brief idea about distance and displacement. Also, I would love to hear your thoughts and feedback about this article via the comments section given below.

And if you find this article useful, don’t forget to share with your friends and colleagues on Facebook and Twitter.

Also, learn the difference between speed and velocity .

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Mechanics (Essentials) - Class 11th

Course: mechanics (essentials) - class 11th   >   unit 4.

• What is displacement?
• Worked example: Properties of Distance and Displacement

Distance and displacement - properties

• Distance and displacement introduction
• Distance and displacement in one dimension
• (Choice A)   9   m ‍   A 9   m ‍  
• (Choice B)   10   m ‍   B 10   m ‍  
• (Choice C)   12   m ‍   C 12   m ‍  

Gurumuda Networks

Distance and displacement – problems and solutions

1. A car travels along a straight road 100 m east then 50 m west. Find distance and displacement of the car.

Distance is 100 met ers + 50 meters = 150 meters

Displacement is 100 meters – 50 meters = 50 meters, to the east .

2. A person walks 4 meters east, then walks 3 meters north. Determine distance and displacement.

3. A runner travels around rectangle track with l ength = 50 meters and width = 20 meters. After travels around rectangle track two times, runner back to starting point. Determine distance and displacement.

Circumference of rectangle = 2(50 meters) + 2(20 meters) = 100 meters + 40 meters = 140 meters.

Travels around rectangle 2 times = 2(140 meters) = 280 meters.

Distance = 280 m .

Displacement = 0 m . (the runner return to the starting point)

4. Car’s speedometer reads 10,500 km at the start of a trip and 10,700 km at the end. Determine distance and displacement.

Distance = 10,700 km – 10,500 km = 200 km.

Displacement = 0 km. (car return to the starting point)

1. What is distance? Answer: Distance is the total movement of an object, irrespective of the direction.

2. What is displacement? Answer: Displacement is the shortest distance between the initial and final position of an object, with a specified direction.

3. Can distance ever be less than displacement? Answer: No. Distance is always greater than or equal to displacement.

4. If a person walks around a circular track and ends up at the starting point, what is the displacement? Answer: The displacement is zero because the initial and final positions are the same.

5. What is the unit of distance and displacement in the metric system? Answer: Both are measured in meters (m).

6. Can displacement be negative? Answer: Yes. A negative displacement indicates the direction is opposite to the chosen positive direction.

7. How are distance and displacement represented in vector form? Answer: Distance is a scalar quantity and has only magnitude. Displacement, being a vector, has both magnitude and direction.

8. If a person walks 10 meters east and then 10 meters west, what is the total distance covered and the total displacement? Answer: Distance = 20 meters; Displacement = 0 meters.

9. What does a zero displacement indicate? Answer: It indicates that the initial and final positions are the same.

10. Is it possible for an object to be in motion if its displacement is zero? Answer: Yes. An object moving in a closed loop or path returns to its starting position, resulting in zero displacement.

11. How is average speed calculated? Answer: Average speed = Total distance traveled / Total time taken.

12. How is average velocity calculated? Answer: Average velocity = Total displacement / Total time taken.

13. What is the significance of the direction in displacement? Answer: The direction in displacement helps in understanding the orientation of movement from the initial to the final position.

14. How do we denote displacement in a one-dimensional motion? Answer: In one-dimensional motion, displacement can be represented by a positive or negative value, depending on the chosen reference direction.

15. What would be the displacement for any object thrown upwards and then coming back to the thrower’s hand? Answer: The displacement would be zero because the starting and ending positions are the same.

16. Can the magnitude of displacement be greater than the distance traveled? Answer: No. The magnitude of displacement can be equal to or less than the distance, but never greater.

17. What role does the reference point play in determining displacement? Answer: The reference point helps define the initial position from which displacement is calculated.

18. If a car travels in a straight line for 50 kilometers and then returns to its starting point, what is its displacement? Answer: The displacement is zero.

19. What is the path length in the context of motion? Answer: Path length is another term for distance traveled.

20. How can you distinguish between distance and displacement graphically? Answer: On a graph, distance is always represented by a non-decreasing curve, while displacement can be represented by a curve that increases, decreases, or remains constant, depending on the direction of movement.

21. Is it necessary for an object to be in motion to have a displacement? Answer: No. An object can have a displacement if its position changes, even if it isn’t in motion for some duration.

23. If the total path covered by an object is curved, is the displacement still the shortest distance between the starting and ending points? Answer: Yes, the displacement is always the straight line distance between the starting and ending points.

24. In which situations can distance and displacement have the same magnitude? Answer: When an object moves in a straight line without changing its direction.

25. Can an object have a constant speed and a changing velocity? Answer: Yes, if the object is moving in a curved path at a constant speed, its direction (and hence velocity) changes, but its speed remains constant.

26. How do distance and displacement relate to time? Answer: Distance and displacement, when divided by time, give average speed and average velocity, respectively.

27. Can an object’s distance from a point increase while its displacement remains constant? Answer: No. If the displacement remains constant, the object’s position relative to the starting point doesn’t change, so the distance from the point cannot increase.

28. Is displacement a scalar or vector quantity? Answer: Displacement is a vector quantity.

29. Why can’t we add distances directly to get a resultant like displacements? Answer: Because distance lacks direction, it represents only magnitude. Displacements have both magnitude and direction, so they can be added vectorially.

30. In which motion is the displacement equal to the circumference of the circle? Answer: In circular motion, if one complete revolution is made, the distance is equal to the circumference, but the displacement is zero. Only if the motion covers exactly half of the circle, the displacement (a straight line through the circle’s diameter) would be equal to the circle’s radius.

• Distance and displacement
• Average speed and average velocity
• Constant velocity
• Constant acceleration
• Free fall motion
• Down motion in free fall
• Up and down motion in free fall

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6.1 Solving Problems with Newton’s Laws

Learning objectives.

By the end of this section, you will be able to:

• Apply problem-solving techniques to solve for quantities in more complex systems of forces
• Use concepts from kinematics to solve problems using Newton’s laws of motion
• Solve more complex equilibrium problems
• Solve more complex acceleration problems
• Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy

Applying newton’s laws of motion.

• Identify the physical principles involved by listing the givens and the quantities to be calculated.
• Sketch the situation, using arrows to represent all forces.
• Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
• Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
• Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 6.2 (a). Then, as in Figure 6.2 (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 6.2 (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure 6.2 (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure 6.2 (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set a y = 0 . a y = 0 . ) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Example 6.1

Different tensions at different angles.

Thus, as you might expect,

This gives us the following relationship:

Note that T 1 T 1 and T 2 T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 T 2 ends up being greater than T 1 T 1 because it is exerted more vertically than T 1 . T 1 .

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for T 2 T 2 in terms of T 1 T 1 reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of T 1 T 1 to be

Finally, we find the magnitude of T 2 T 2 by using the relationship between them, T 2 = 1.225 T 1 T 2 = 1.225 T 1 , found above. Thus we obtain

Significance

Particle acceleration.

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Example 6.2

Drag force on a barge.

The drag of the water F → D F → D is in the direction opposite to the direction of motion of the boat; this force thus works against F → app , F → app , as shown in the free-body diagram in Figure 6.4 (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x - and y -axes are in the same direction as F → 1 F → 1 and F → 2 . F → 2 . The problem quickly becomes a one-dimensional problem along the direction of F → app F → app , since friction is in the direction opposite to F → app . F → app . Our strategy is to find the magnitude and direction of the net applied force F → app F → app and then apply Newton’s second law to solve for the drag force F → D . F → D .

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that F → D F → D is in the opposite direction of F → app , F → app , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F → app , F → app , but its magnitude is slightly less than F → app . F → app . The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water F D F D in terms of known quantities:

Substituting known values gives

The direction of F → D F → D has already been determined to be in the direction opposite to F → app , F → app , or at an angle of 53 ° 53 ° south of west.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

Example 6.3

What does the bathroom scale read in an elevator.

From the free-body diagram, we see that F → net = F → s − w → , F → net = F → s − w → , so we have

Solving for F s F s gives us an equation with only one unknown:

or, because w = m g , w = m g , simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes F s − w = − m a . F s − w = − m a . )

• We have a = 1.20 m/s 2 , a = 1.20 m/s 2 , so that F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) yielding F s = 825 N . F s = 825 N .
• Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because a = Δ v Δ t a = Δ v Δ t and Δ v = 0 . Δ v = 0 . Thus, F s = m a + m g = 0 + m g F s = m a + m g = 0 + m g or F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , which gives F s = 735 N . F s = 735 N .

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure 6.5 (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding 6.1

Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20 m/s 2 . 1.20 m/s 2 .

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Example 6.4

Two attached blocks.

For block 1: T → + w → 1 + N → = m 1 a → 1 T → + w → 1 + N → = m 1 a → 1

For block 2: T → + w → 2 = m 2 a → 2 . T → + w → 2 = m 2 a → 2 .

Notice that T → T → is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

When block 1 moves to the right, block 2 travels an equal distance downward; thus, a 1 x = − a 2 y . a 1 x = − a 2 y . Writing the common acceleration of the blocks as a = a 1 x = − a 2 y , a = a 1 x = − a 2 y , we now have

From these two equations, we can express a and T in terms of the masses m 1 and m 2 , and g : m 1 and m 2 , and g :

Check Your Understanding 6.2

Calculate the acceleration of the system, and the tension in the string, when the masses are m 1 = 5.00 kg m 1 = 5.00 kg and m 2 = 3.00 kg . m 2 = 3.00 kg .

Example 6.5

Atwood machine.

• We have For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . (The negative sign in front of m 2 a m 2 a indicates that m 2 m 2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . Solving for a : a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 . a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 .
• Observing the first block, we see that T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N . T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N .

Check Your Understanding 6.3

Determine a general formula in terms of m 1 , m 2 m 1 , m 2 and g for calculating the tension in the string for the Atwood machine shown above.

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6

What force must a soccer player exert to reach top speed.

• We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ v = 8.00 m/s Δ v = 8.00 m/s . We are given the elapsed time, so Δ t = 2.50 s . Δ t = 2.50 s . The unknown is acceleration, which can be found from its definition: a = Δ v Δ t . a = Δ v Δ t . Substituting the known values yields a = 8.00 m/s 2.50 s = 3.20 m/s 2 . a = 8.00 m/s 2.50 s = 3.20 m/s 2 .
• Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, F net = m a . F net = m a . Substituting the known values of m and a gives F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N . F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N .

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Check Your Understanding 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7

What force acts on a model helicopter.

The magnitude of the force is now easily found:

Check Your Understanding 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8

Baggage tractor.

• ∑ F x = m system a x ∑ F x = m system a x and ∑ F x = 820.0 t , ∑ F x = 820.0 t , so 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . Since acceleration is a function of time, we can determine the velocity of the tractor by using a = d v d t a = d v d t with the initial condition that v 0 = 0 v 0 = 0 at t = 0 . t = 0 . We integrate from t = 0 t = 0 to t = 3 : t = 3 : d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s . d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s .
• Refer to the free-body diagram in Figure 6.8 (b). ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N . ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N .

Recall that v = d s d t v = d s d t and a = d v d t a = d v d t . If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have d t = d s v d t = d s v and d t = d v a . d t = d v a . Now, equating these expressions, we have d s v = d v a . d s v = d v a . We can rearrange this to obtain a d s = v d v . a d s = v d v .

Example 6.9

Motion of a projectile fired vertically.

The acceleration depends on v and is therefore variable. Since a = f ( v ) , a = f ( v ) , we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, h = 114 m . h = 114 m .

Check Your Understanding 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Interactive

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

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Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction

• Authors: William Moebs, Samuel J. Ling, Jeff Sanny
• Publisher/website: OpenStax
• Book title: University Physics Volume 1
• Publication date: Sep 19, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
• Section URL: https://openstax.org/books/university-physics-volume-1/pages/6-1-solving-problems-with-newtons-laws

© Jan 19, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

• (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20