Question

A particle starts from rest at time t = 0 and moves in a straight line with an

acceleration a (m s )

-2

. Find the time at which the (a) speed of the particle will be

4 times its speed at t = 2 s and (b) distance covered by it will be 8 times that

travelled by the particle at t = 2 s​

Answer 1

v=u+at

initially the particle is at rest so u=0

therefore velocity at t=2 is v=a*2

1) now initial velocity is 2a and final velocity is 8a

so for time,

u=0

8a=0+at

therefore t=8

2) apply formula s=ut+(1/2)a*t^2

distance travelled at t=2s

u=0

s=(1/2)a*4

therefore s=2a

so the time taken for the body to travel 8 times the distance travelled at t=2s is

s=ut + (1/2)a*t^2

u=0

so,

16a= (1/2)a*t^2

t^2=32

therefore t=5.66