[Chapter 2 Objectives]

BHS -> Staff -> Mr. Stanbrough -> Physics -> Mechanics -> Kinematics -> this page

An airplane, initially moving at 3.0 m/s down a runway, begins to accelerate down the runway at 3.6 m/s^{2}. How far down the runway will it be in 20 seconds?

v

_{o}= 3.0 m/sa = 3.6 m/s

^{2}= 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/s

^{2}. 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/s^{2}, 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:

v

_{o}= 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 v

_{o}vanishes, since v_{o}= 0 m/s):Substituting values:

Answer:The acceleration of the car is about 21 m/s^{2}.This is about 2g's (g = 0.8 m/s

^{2}) - 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/s^{2}. 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 g_{moon}= 1.6 m/s^{2}?

= 20 m

v

_{o}= 0 m/sa = g

_{moon}= 1.6 m/s^{2}= ?

The kinematics equation that relates , vo, a, and is:

Solving for (and keeping in mind that the term containing v

_{o}vanishes, since v_{o}= 0 m/s):substituting

Answer:It would take the rock about 4.0 seconds to fall.

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 g_{moon}= 1.6 m/s^{2}) if it starts with a downward velocity of 2.0 m/s?

= 20 m

v

_{o}= 2.0 m/sa = g

_{moon}= 1.6 m/s^{2}= ?

The kinematics equation that relates , vo, a, and is:

Although this problem looks a lot like example 7, the v

_{o}t 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