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 45.1m 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 38.3m from where it began decelerating. The entire trip from start to finish (starting and ending at rest) lasts for a duration of 18.4 s.

a) What is the duration of the first part of the trip (the acceleration phase)?
b) What is the duration of the second part of the trip (the deceleration phase)?
c) What is the car's average speed over the course of the entire trip?
d) What is the magnitude of the initial rate of acceleration, a1?
e) What is the magnitude of the subsequent rate of deceleration, a2?

Please add formulas

Explanation / Answer

a1 t12 = 90.2                      eq-1

similarly

a2 t22 = 76.6                     eq-2

also , a1 t1 = a2 t2

a1/a2 = t2/t1    eq-3

dividing eq-1 by eq-2

(a1/a2) (t1/t2)2 = 90.2/76.6

using eq-3

(t2/t1) (t1/t2)2 = 90.2/76.6

t1/t2 = 90.2/76.6

also t1 + t2 = 18.4

hence

(18.4 - t2)/t2 = 90.2/76.6

t2 = 8.45 sec

t1 = 18.4 - 8.45 = 9.95 sec

average speed = total distance / total time = (45.1 + 38.3)/18.4 = 4.53 m/s

from eq-1

a1 t12 = 90.2  

a1 (9.95)2 = 90.2  

a1 = 0.91 m/s2

from eq-2

a2 t22 = 76.6  

a2 (8.45)2 = 76.6  

a2 = 1.1 m/s2

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