A car starts from rest, and begins accelerating at a constant rate a1. It accele

Question

A car starts from rest, and begins accelerating at a constant rate a1. It accelerates at this rate for a distance of 78.0m from its starting point and then immediately begins to decelerate at a different constant rate a2 eventually coming to rest again after traveling an additional distance of 56.5m from where it began decelerating. The entire trip from start to finish (starting and ending at rest) lasts for a duration of 15.1s.

a) What is the magnitude of the initial rate of acceleration, a1?

b) What is the magnitude of the subsequent rate of deceleration, a2?

c) What is the duration of the first part of the trip (the acceleration phase)?

d) What is the duration of the second part of the trip (the deceleration phase)?

e) What is the car's average speed over the course of the entire trip?

Explanation / Answer

Here, 78 = (v/2) * t1

also., 56.5 = (v/2) * t2

And, t1 + t2 = 15.1 sec

=> t1 = 8.76 sec    and t2   = 6.34 sec

a)   magnitude of the initial rate of acceleration = 78*2/8.76*8.76

                                                                           = 2.033 m/sec2

b)   magnitude of the subsequent rate of deceleration = 2.033 * 8.76/6.34

                                                                                      = 2.809 m/sec2

c)   duration of the first part of the trip = 8.76 sec

d)   duration of the second part of the trip   = 6.34 sec

e)     car's average speed over the course = (78+56.5)/(8.76+6.34)

                                                                      = 8.907 m/sec

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