a car starts from the rest and begins accelerating at a constant rate a1. it acc

Question

a car starts from the rest and begins accelerating at a constant rate a1. it accelerates at this rate for a distance of 49.9 meters from its starting point and then immediately begins to decelerate at a different constant rate a2. eventually coming to rest again after traveling a additional distance of 30.4 meters from where it began decelerating. The entire trip from start to finish last for a duration of 24.9 seconds.
A) what is the duration of the first part of the trip?
B) What is the duration of the second part of the trip?.
C) what is the cars average speed over the course of the entire trip?
D) what is the magnitude of the initial rate of acceleration?
E) what is the magnitude of the subsequent rate of deceleration?

Explanation / Answer

A)

x1 = vo*t + (1/2)*a1*t1^2

49.9 = (1/2)*a1*t1^2

2*49.9 = a1*t1^2

a1*t1^2 = 99.8 ........(1)

after t1 time

v1 = a1*t1

t1 + t2 = 24.9

t2 = 24.9 - t1

for second part of journey

x2 = (v1+v2)/2*t2

30.4 = a1*t1*(29.4-t1)/2.............(2)

solving 1 & 2

time t1 = 18.3 s <<<<<=======answer

-----------------

(B)

t2 = 24.9 - t1 = 6.6 s <<<<<---------answer

------------------

(C)

average speed = total distance /total time = (49.9+30.4)/24.9 = 3.22 m/s

===============

(D)

from (1)

a1 = 99.8/18.3^2

a1 = 0.298 m/s^2

---------------------------

part(E)

a2 = v1/t2 = a1*t1/t2

a2 = 0.298*18.3/6.6

a2 = 0.826 m/s^2 <<<<------answer

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