A particle starting from rest moves with constant acceleration. If it takes 5.0 s to reash the speed 18.0 km/h find a. the average velocity during this period, and b. the distance travelled by the particle during this period.

Answer

A, B

Explanation

Given, `u=0
v=18km/hr
=5m/s
t=5 sec
a=v-ut
=5-0xx5
=5m/s^2
S=ut+1/2at^2ltbrge1/2xx1xx(5xx5)
=12.5m
a. Average velocity
V_(ve)=(12.5)/5
=2.5 m/s
b. Distance travelled = 12.5 m`.