problem solving with kinematics equation 1

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Kinematic Equations and Problem-Solving

• Kinematic Equations Introduction
• Solving Problems with Kinematic Equations
• Kinematic Equations and Free Fall
• Sample Problems and Solutions
• Kinematic Equations and Kinematic Graphs

The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6 . The four kinematic equations are:

In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in v i ) indicates that the velocity value is the initial velocity value and a subscript of f (as in v f ) indicates that the velocity value is the final velocity value.

Problem-Solving Strategy

• Construct an informative diagram of the physical situation.
• Identify and list the given information in variable form.
• Identify and list the unknown information in variable form.
• Identify and list the equation that will be used to determine unknown information from known information.
• Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information.
• Check your answer to insure that it is reasonable and mathematically correct.

The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below.  

Example Problem A

Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s 2 , then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.)

The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the v f value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity ( v i ) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration ( a ) of the car is given as - 8.00 m/s 2 . (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below.  

The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are v f , v i , a , and d . Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables.

Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.

(0 m/s) 2 = (30.0 m/s) 2 + 2 • (-8.00 m/s 2 ) • d

0 m 2 /s 2 = 900 m 2 /s 2 + (-16.0 m/s 2 ) • d

(16.0 m/s 2 ) • d = 900 m 2 /s 2 - 0 m 2 /s 2

(16.0 m/s 2 )*d = 900 m 2 /s 2

d = (900 m 2 /s 2 )/ (16.0 m/s 2 )

The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!  

Example Problem B

Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s 2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period.

Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the v i value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration ( a ) of the car is 6.00 m/s 2 . And the time ( t ) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below.

The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, v i , a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables.  

d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s 2 ) • (4.10 s) 2

d = (0 m) + ½ • (6.00 m/s 2 ) • (16.81 s 2 )

d = 0 m + 50.43 m

The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6 , we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions.  

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Sample Kinematics Problems with Solutions

Following are a variety of problems involving uniformly accelerated motion along a line.  In the solution a list of known quantities will be given followed by a list of quantities wanted.  The equations to be used will be identified by number from the list below, but the algebraic work of solving the equations will be left to the student.  In most of the problems there will be a discussion of alternate methods of solution or suggestions and hints as to how to attack this particular type of problem.  Do not try to just read this section.  You must use pencil, paper and a calculator to obtain the maximum value from studying these problems.

Sample Problem #1

A car starts from rest and accelerates with a uniform acceleration of 10 ft s 2 .  Calculate how long it will take for the car to reach a speed of 90 ft s (slightly over 60 mph), and calculate the distance the car moves during this time.

Sample Solution #1        

Given V i 0 a 10 ft s 2 v f 90 ft s

Needed ΔX Δt

This problem can be solved by a straight forward application of equation 4 to find Δt and equation 5 to find ΔX.   The student should demonstrate that the correct answers are Δt=9 s and ΔX=405 ft.  It is a good idea to try to find the answers  using slightly different procedures whenever possible.  For example, in this problem a quick check can be made by using your calculated value for Δt, from equation 4 to find the average speed and then set X=V avg Δt to get 405 ft a different way.  A second different method would be to use equation 6 and the value of Δt obtained from equation 4 to calculate ΔX.  However, while both of the alternate techniques are good checks on the validity of your calculations it is best to avoid using an answer to one part as a basis for a second  calculation whenever possible.

Sample Problem #2

A car moving at 30 m s stops with a constant acceleration in a distance of 100 m.  Calculate the acceleration and the time to stop.

Sample Solution #2

Given V i 30 m s V f 0 m s Δx 100 m

Needed Δt a

To solve the problem apply equation 5 to find the acceleration (a=-4.5 m s 2 ).  Use the calculated value of the acceleration to find the time (Δt=6.67 s) using either equation 4 or 6 (4 is easier).

Sample Problem #3

Explain the significance of the negative sign in problem 2.

Sample Solution #3

The acceleration is in the negative direction.  Since V was arbitrarily assigned a positive direction, the acceleration must be in the opposite direction.

Sample Problem #4

Assume that the car in problem 2 keeps the same acceleration for an additional 5 seconds.  Find its speed and position at the end of this time.

Sample Solution #4

Given V i 0 a -4.5 m s 2 Δt 5s

Needed V f X

Notice that here we are not asked for ΔX but are asked for X, the position at the end of the 5th second.  We need to be sure to specify the position unambiguously,  either with reference to the car's position at the start of problem 2 or with respect to its position at the end of problem 2 and the start of this problem.  We will place the origin of the reference system at the position of the car at the start of problem 2.  Thus at the end of problem 2 it has a position of X=100 m.    Using equation 6 with the data listed above we see that ΔX= -56.25 m.  The final position of the car is therefore 100 - 56.25 or 43.75 m.    To find the speed use equation 4 and obtain a speed of  - 22.5 m s .  The complete answer to the problem could be stated as follows: "The car is 43.75 m from its starting point in the direction it was originally moving. It is moving back toward the starting point at a speed of 22.5 m s and is accelerating toward the starting point with an acceleration of 4.5 m s 2 ."  This statement is much more meaningful than simply writing X = 43.75 m and V = -22.5 m s

Sample Problem #5

 A baseball is batted vertically with an initial speed of 45 m s .  Calculate:     (a)     the time before it returns to the level at which it started.     (b)     the maximum height the ball reaches.     (c)     the position, speed and acceleration 6 seconds after it is batted.     (d)     the position and speed 12 seconds after it is batted.

Sample Solution #5

Read this problem carefully and decide the order that you will do the problem. Keep in mind the facts mentioned in the hints such as: the time to reach the top is equal to one-half the total time in the air; the speed at the top is zero; and the speed when it reaches the same level is the same as the initial speed.  In multi part problems such as this one, it will be necessary to list the given quantities for each part.  As you solve the problem, check to be sure the answers to the various sections are consistent with each other.  Before starting the problem decide which direction to call positive.  In this problem we will call the upward direction positive and the downward direction negative.

Sample Problem #6

A ball dropped from the roof of a tall building passed a window ledge with a speed of 96 ft s and struck the ground 1.0 s later.     (a)     What is the height of the window?     (b)     How tall is the building?

Sample Solution #6   

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