An Alternate Solution for Example 5


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Example 5:

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:

Example 5 diagram

vo = 3.0 m/s

a = 3.6 m/s2

delta t = 20 s

delta x = ?

The plan: We can use delta_x = v_bar*delta_t to find the distance the plane moves - except that we don't know the plane's average velocity, v_bar. We could find the plane's average velocity using the equation v_bar = (vo + v)/2, except we don't know the airplane's final velocity, v. Well, we could find the plane's final velocity from the equation v = vo + a*delta_t. Ha! This is a three-step, round-about solution, but it is a valid solution. It shows that sometimes you just have to get started, and find whatever you can find - then work from there.

delta_x = v_bar*delta_t

v_bar = (v + vo)/2
v = vo + a*delta_t

v = 75 m/s

v_bar = 39 m/s

delta_x = 780 m

Answer: The airplane moves 780 meters in this time.


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last update August 5, 2001 by JL Stanbrough