Question

a car starts from rest and moves with constant acceleration of 4 m/s² for 30 seconds .then 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 of car.
3.total distance covered by car.
4. average speed of car through out of motion.
please it's urgent​

Answer 1

for first part of motion

here u=0;

a=4m/s²;

t=30sec

then

v=u+at

v=0+4*30

v=120m/s (this is maximum velocity)

when break apply car come to rest but according to question acceleration of car is constant I.e. 4m/s²

distance for first part of motion

here u=0;

a=4m/s²;

t=30sec;

s=ut+1/2at²

s=0*30+1/2(4)(30)²

s=1800m or 1.8km

for second part of motion

here initial velocity of car is 120m/s and final velocity is zero

so

u=120m/s;

v=0m/s;

t=30+60=90sec;

a=?

v=u+at

a=(v-u) /t

a=(0-120)/90

a=-120/90

a=-1.33m/s² (retardation effect)

distance for second part of motion

s=ut+1/2at²

s=120*90+1/2(-1.33)(90)²

s=10800+(-5386.5)

s=5413.5m

total distance cover

1800+5413.5=7213.5m

since average velocity of the car is

V=1/2(V+U)

V=1/2(0+120)

V=60m/s