Question

a car starts from rest and moves with constant acceleration of 4 m per second for 30 seconds.Then the brakes are applied and the car comes to rest in another 60 seconds.Find-
1)Maximum velocity attained by the car.
2)Magnitude of retardation.
3)Total distance covered by the car.
4)Average speed of the car through out the motion.

Answer 1

Answer:

The car is undergoing acceleration for first 30 seconds and deceleration for next 60 seconds.

consider first case:

u=0,a=4m/sec²,t=30 sec

here v=?

v=u+at

v=0+4*30=120m/sec

which is max velocity gained by car

In the second case

Initial velocity,u=120m/sec

t=60sec

v=0

from,v=u+at

0=120+a*60

a=-2m/sec²

Now,Total distance covered=Distance in first case+Distance in second case

s=S1+S2

=(ut+½at²)+(ut+½at²)

=(0*30+½*4*900)+(120*60+½*(-2)*3600)

=5430metres

finally Average speed is Total distance /Total time

AVG speed=5430/(30+60)

=5430/90

=60.3m/sec