How to Solve Kinematic Equations – Kinematics is the branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without consideration of the causes of motion. There are four kinematic equations when the initial starting position is the origin, and the acceleration is constant:
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 vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value.
Table of Contents
Problem-Solving Strategy
In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object’s motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps:
- 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/s2, 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 vf value can be inferred to be 0 m/s since Ima’s car comes to a stop. The initial velocity (vi) 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/s2. (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.
| Diagram: | Given: | Find: |
|---|---|---|
 | vi = +30.0 m/s vf = 0 m/s a = – 8.00 m/s2 | d = ?? |
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 vf, vi, 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. vf2 = vi2 + 2 • a • d
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/s2) • d
0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d
(16.0 m/s2) • d = 900 m2/s2 – 0 m2/s2
(16.0 m/s2)*d = 900 m2/s2
d = (900 m2/s2)/ (16.0 m/s2)
d = (900 m2/s2)/ (16.0 m/s2)
d = 56.3 m
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/s2 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 vi 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/s2. 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.
| Diagram: | Given: | Find: |
|---|---|---|
 | vi = 0 m/s t = 4.10 s a = 6.00 m/s2 | d = ?? |
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, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables.
d = vi • t + ½ • a • t2
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.
d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2
d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2)
d = 0 m + 50.43 m
d = 50.4 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.
Avoiding Common Mistakes: Top of Flight
A special example of the hidden quantity is when they tell you an object is at “the top of its flight/motion/path/etc.” This means they are secretly telling you that vf is 0 because an object traveling in one dimension always has a velocity of 0 at the top of its path. Wondering why? Well, if the velocity was going up, then a millisecond later, the object would be higher (and thus it can’t be at the top of its path). Likewise, if the object had a velocity downward, then it would have been higher a millisecond before (so it can’t be at the top either).
Avoiding Common Mistakes: Positive and Negative Numbers
It can be tricky to keep track of your negatives. The key is direction; down is always negative. So if an object is going down, it will have a negative velocity. If the acceleration is going down (which it almost always is) then the acceleration is negative. And don’t forget from our first chapter, it is possible to have positive velocity and negative acceleration at the same time!
Conclusion
The kinematic equations are a set of equations that describe the motion of an object with constant acceleration. Kinematics equations require knowledge of derivatives, rate of change, and integrals. To keep our focus on high school physics, we will not be covering integrals.