Question

a body has initially velocity of 36km/hr and moves under a constant retardation (-a)of 2m/s² calculate.(1)its velocity after 3 sec.(2)the time after which it will stop.(3)the distance it will travel before coming to rest.

Answer 1

velocity after 3 sec = 30m/s
time required to stop = 18s
total distance traveled=324m

Answer 2

u = 10 m/s
a = -2m/s^2
t= 3s
1.) v = u+at
= 10+(-2)3
= 4m/s^2
2.) v=0
v= u+at
0=10+(-2)t
t= 5 s
3.) v^2 - u^2 =2as
0 - 100 =2×(-2)s
s= 25 m