Question
A sports car moving at constant speed travels 120 m in 5.1 s. If it then brakes and comes to a stop in 4.0 s, what is its acceleration? Express the answer in terms of g's, where g = 9.80 m/s2. (Take the positive direction to be the direction of travel.)
Answers
Initial speed before braking = 120/5.1 = 23.53 m/s
Braking acceleration rate = 23.53 m/s/4.0 is (Vfinal -Vinitial)/(time)
= -5.88 m/s^2
Divide that by g = 9.81 m/s^2 for the number g's.
Braking acceleration rate = 23.53 m/s/4.0 is (Vfinal -Vinitial)/(time)
= -5.88 m/s^2
Divide that by g = 9.81 m/s^2 for the number g's.
part a. 5.88 m/s^2
part b. .6g
part b. .6g
a car moving constants speed travels 150 m in .8.0 s. it then brakes and comes to a stop after 5.0. Determine its deceleration
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