Practice Calculating Motion Using Kinematic Equations & Graphs

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  • 0:03 The Kinematics Toolbox
  • 1:46 Example 1 & 2
  • 3:46 Example 3 & 4
  • 5:43 Example 5 & 6
  • 7:08 Lesson Summary
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Lesson Transcript
Instructor: Matthew Bergstresser

Matthew has a Master of Arts degree in Physics Education. He has taught high school chemistry and physics for 14 years.

Motion has many components associated with it. This lesson will tie all of the components of motion together by using the kinematics equations and graphs of motion.

The Kinematics Toolbox

Carpenters have toolboxes with specialty tools they use to build houses. A mechanic's toolbox contains tools to turn bolts in car engines, among other things. Physicists have a toolbox that contains many equations. When a physics problem arises, they pull out the five kinematic equations and use the equations that are appropriate for the problem they are solving. The following are the five kinematics equations:


Notice that there are arrows over all of the variables that are vectors. This means we have to be very careful with labeling which directions are positive.

Another tool at our disposal is a graph interpretation chart. This tells us what the slope and area between the curve and x-axis mean on each type of a kinematics graph.


We have a few problems that involve only using kinematics equations, and some that involve graphs. Let's grab our toolbox and get busy!

Example 1 & 2

Prompt: How long does it take a rock dropped from rest to fall 320 meters from the bridge into a river? Ignore air resistance.

Solution: The first step in solving any problem is to gather the data, isolate what is being asked, and draw a simple sketch of the scenario:

v0 = 0 m/s

d = 320 m

a = g = 9.8 m/s2


We can use Equation 3 to solve this problem. First, we will solve it for t:


Now we can plug in the data:


The answer is t is approximately 8.1 seconds.

Now, let's look at Example 2:

Prompt: A drone starts on the ground at rest, and moves straight up to an elevation of 100 meters in 13 seconds. What is the acceleration of the drone?

Solution: We start with organizing the data:

v0 = 0 m/s

d = 100 m

t = 13 seconds

a = ?

Now we draw a sketch:


Now we can choose which kinematics equation will get us to the answer the simplest way possible. Let's use Equation 3:


Plugging in our values for displacement and time, the solution is a is approximately 1.2 meters per second squared.

Let's check out Example 3.

Example 3 & 4

Prompt: A bird has an initial horizontal velocity of 5 m/s. It covers 16 meters in 20 seconds. What is its velocity at 20 seconds?

Solution: Listing the data we get:

v0 = 5 m/s

d = 16 m

t = 20 s

vf = ?


This scenario requires two kinematics equations. We will use Equations 2 and 5.


We will solve the equation on the left for a, and plug it into the equation on the right:


This looks like a quadratic equation because v and v2 are in it. We rearrange this equation so that we can use the quadratic equation and get:


Using the quadratic equation to solve for v, we get v equals 5 meters per second and -3.4 meters per second.

Consider the solution 5 meters per second. The kinematic equations only apply when acceleration is constant. So if the velocity of the bird at twenty seconds was +5 meters per second, and the velocity of the bird at zero seconds was +5 meters per second, then we know that the bird had a constant velocity of +5 meters per second throughout the trip. Since distance equals velocity times time, the distance covered in twenty seconds at a pace of 5 meters per second is:

d = v * t = (5 m/s)*(20 s) = 100 m ≠ 16 m

Therefore, we know that the final velocity must be equal to -3.4 meters per second. Physically, this means that the bird probably flew in an arc to turn around from the positive direction to the negative direction.

Let's check out Example 4:

Prompt: Use Graph 1 to determine the displacement between 0 and 3 seconds.


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