Question

8. A car starts from rest and moves with constant
acceleration of 4 m/s2 for 30 seconds. Then the
brakes are applied and the car comes to rest in
another 60 seconds. Find [5]
(i) Maximum velocity attained by car
(ii) Magnitude of retardation of car
(iii) Total distance covered by car
(iv) Average speed of car through out the motion

Answer 1

Explanation:

(i) Given

Initial velocity (u)=0

acceleration (a)=4m/s^2

time (t)=30s

It is given that the car accelerates and then retards.so, the maximum velocity attained by the car will be the final velocity of its acceleration and Initial velocity for its retardation

v=u+at

v=0+4×30

v=120m/s

maximum velocity attained by the car is 120m/s

(ii) Here

initial velocity (u)=120m/s

final velocity(v)=0

time=60s

retardation=v-u/t

=-120/60

=-2m/s^2

magnitude of retardation of car is 2m/s^2

(iii) distance covered when it accelerates

s=ut+1/2 at^2

S1= 1/2 ×4×30×30

=1800m

distance covered when it retards

S2=120×60+1/2 (-2)×60×60

=3600m

total distance=S1+S2=5400m

(iv) total distance=5400m

total time=90s

average speed=total distance/total time

=5400/90

=60m/s