A cameraman on a pickup truck is traveling westward at 25 km/h while he videotap

Question

A cameraman on a pickup truck is traveling westward at 25 km/h while he videotapes a cheetah that is moving westward 28 km/h faster than the truck. Suddenly, the cheetah stops, turns, and then run at 41 km/h eastward, as measured by a suddenly nervous crew member who stands alongside the cheetah's path. The change in the animal's velocity takes 2.2 s. What are the (a) magnitude and (b) direction of the animal’s acceleration according to the cameraman and the (c) magnitude and (d) direction according to the nervous crew member?

Explanation / Answer

acceleration = v/t Now assuming westward is positive, the cameramen sees an initail velocity of 28km/h*(1000m/km*1h/3600s) = 7.778m/s and
a final velocity of -(41+25)*1000/3600 = -18.33 m/s

So for the cameraman a = (-18.33- 7.778)/2.0 = -13.054 m/s^2 so magnitude = 13.054m/s^2

For the stationary crew member vi = 53km/h*(1000/3600) = 14.72 m/s
and a vf = -41*1000/3600 = -11.388m/s

So he sees an a = (-11.388 -14.72)/2.0s = -13.054 m/s^2 (magnitude 13.054m/s^2)

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