# Using the Kinematics Equation      [Chapter 2 Objectives]

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### Example 5 - Finding displacement:

An airplane, initially moving at 3.0 m/s down a runway, begins to accelerate down the runway at 3.6 m/s2. How far down the runway will it be in 20 seconds?

### Solution: vo = 3.0 m/s a = 3.6 m/s2 = 20 s = ?

The kinematics equation that relates , vo, a, and is: (Click here to see a derivation.)

substituting: (Note: A very common blunder is to forget to square the clock reading in the second term!)

Answer: The airplane will be 780 m down the runway in 20 seconds.

This seems like a long way - is it a reasonable answer? Well, suppose that the airplane accelerates at 3 m/s2. In 20 seconds, its velocity will increase by 60 m/s (3*20), so its speedometer will say about 63 m/s. Its average velocity during this time will be about 33 m/s ( = (3 m/s + 63 m/s)/2). If you have an average velocity of 33 m/s for 20 seconds, you will travel about 660 m (distance = average velocity times time.) If the plane's acceleration were 4 m/s2, It's velocity would have increased 4*20 = 80 m/s in 20 seconds, so it would be going 83 m/s, and its average velocity during this time would be (3 m/s + 83 m/s)/2 = 86 m/s/2 = 43 m/s. Since distance equals average velocity times time, it would have traveled 20*43 meters, which is over 800 meters. Yes, our answer is reasonable.

Here is an alternate solution to this problem based on this argument.

### Example 6 - Finding acceleration:

A car, initially at rest, begins to accelerate uniformly. If the car travels 1600 m in 12.4 seconds, what was the car's acceleration?

### Solution: vo = 0.0 m/s = 1600 m = 12.4 s a = ?

The kinematics equation that relates , vo, a, and is: Solving for a (and keeping in mind that the term containing vo vanishes, since vo = 0 m/s): Substituting values: This is about 2g's (g = 0.8 m/s2) - is it a reasonable answer? Let's see. Suppose it took 10 seconds to cover the 1600 meters. Since , the car's average velocity would need to be about 160 m/s to cover 1600 meters in 10 seconds (1600 m/10 s). Now, , the final velocity of the car would have to be twice the average velocity - about 320 m/s. Now, , the acceleration would be 320 m/s / 10 s = 32 m/s2. I think our answer is at least "in the ballpark".

### Example 7 - Finding time (when original velocity = 0):

How long would it take a rock to fall 20 meters from rest on the Moon, where gmoon = 1.6 m/s2?

### Solution:  = 20 m vo = 0 m/s a = gmoon = 1.6 m/s2 = ?

The kinematics equation that relates , vo, a, and is: Solving for (and keeping in mind that the term containing vo vanishes, since vo = 0 m/s): substituting ### Example 8 - Finding time (when original velocity is not 0):

How long would it take a rock to fall 20 meters on the Moon (where gmoon = 1.6 m/s2) if it starts with a downward velocity of 2.0 m/s?

### Solution:  = 20 m vo = 2.0 m/s a = gmoon = 1.6 m/s2 = ?

The kinematics equation that relates , vo, a, and is: Although this problem looks a lot like example 7, the vot term does not disappear, which makes this a whole different ball game. What we have is a quadratic equation in . In some situations you may be able to solve this equation by factoring, but, more likely, you will have to resort to using the quadratic formula: Substituting and rearranging: Comparing this equation to the standard-form quadratic shown above, we see that: substituting into the quadratic formula gives: Answer: It would take the rock about 3.9 seconds to fall. (Can you figure out what the -6.41 s represents?)     [Chapter 2 Objectives]

BHS -> Staff -> Mr. Stanbrough -> Physics -> Mechanics -> Kinematics -> this page
last update August 23, 2004 by JL Stanbrough