A car starts from rest, and begins accelerating at a constant rate a 1 . It acce

Question

A car starts from rest, and begins accelerating at a constant rate a1. It accelerates at this rate for a distance of 78.2 m from its starting point and then immediately begins to decelerate at a different constant rate a2eventually coming to rest again after traveling an additional distance of 46.4 m from where it began decelerating. The entire trip from start to finish (starting and ending at rest) lasts for a duration of 19.3 s.

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

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

c) What is the car's average speed over the course of the entire trip? ____________m/s

d) What is the magnitude of the initial rate of acceleration, a1? ________m/s2

e) What is the magnitude of the subsequent rate of deceleration, a2? ________m/s2

Explanation / Answer

plot of velocity-time graph is a triangle with height=maximum velocity achieved and base =total time duration

as distance travelled=area under velocity-time plot

==>78.2+46.4=0.5*maximum velocity*19.3

==>maximum velocity=(78.2+46.4)/(0.5*19.3)=12.912 m/s

part a:

for the first part:
initial speed=0

final speed=12.912 m/s

distance=78.2 m

then acceleration=(final speed^2-initial speed^2)/(2*distance)

=12.912^2/(2*78.2)

=1.066 m/s^2

time taken=(final speed-initial speed)/acceleration

=12.912/1.066=12.133 seconds

part b:

time taken for second phase=19.3-12.133=7.187 seconds

part c:

average speed=total distance/total time

=(78.2+46.4)/19.3

=6.456 m/s

part d:

as derived in part a, initial accleration=1.066 m/s^2

part e:

second deceleration=(final speed-initial speed)/time

=(0-12.912)/7.187

=-1.7966 m/s^2

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