9 November, 05:47

# A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 28.0 m/s. Then the vehicle moves for 41.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.(a) How long is the ride-sharing car in motion (in s) ?(b) What is the average velocity of the ride-sharing car for the motion described? (Enter the magnitude in m/s.)

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1. 9 November, 06:46
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Time taken to accelerate to 28 m / s

= 28 / 2 = 14 s

a) Total length of time in motion

= 14 + 41 + 5

= 60 s.

b)

Distance covered while accelerating

s = ut + 1/2 at²

= 0 +.5 x 2 x 14²

= 196 m.

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

distance covered while decelerating

v = u - at

0 = 28 - a x 5

a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m.

Total distance covered

= 196 + 1148 + 70

= 1414 m

total time taken = 60 s

average velocity

= 1414 / 60

= 23.56 m / s.