A point initially at rest moves along x-axis. Its acceleration varies with time as `a = (6t + 5) m//s^(2)`. If it starts from origin, the distance covered in 2 s is:

A point mass initially at rest moves along x-axis. Its acceleration varies with time as a = 6t+ 5 ms. If it starts from origin, the distance covered by it in 1 s is

A point initially at rest moves along x axis. It's acceleration varies with time as a=(6t+5) in m//s^2 . If it starts from the origin, what is the distance covered in 2 seconds.

A particle initially at rest starts moving along x-axis.Its acceleration varies with time as a=6tm/s^(2) .If it starts from x=10m .Then the position of particle at t=2s is

Acceleration of a particle varies with time as a=12t^(2) m/ s^(2) . If the particle starts its motion from rest,then its displacement in initial 2 seconds will be ?

A particle starts from origin at time t=0 and moves in positive x-direction. Its velocity v varies with times as v = 10that(i) cm/s. The distance covered by the particle in 8 s will be

A particle starts from rest. Its acceleration is varying with time as shown in the figure. When the particle comes to rest, its distance from its starting point is

A particle starts from rest. Its acceleration is varying with time as shown in the figure. When the particle comes to rest, its distance from its starting point is

A particle starts from rest and moves with acceleration a which varies with time t as a=kt where k is a costant. The displacement s of the particle at time t is