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 ?

If displacements of a particle varies with time t as s = 1/t^(2) , then.

The acceleration a of a particle in m/s^2 is give by a=18t^2+36t+120 . If the particle starts from rest, then velocity at end of 1 s is

The acceleration of a particle varies with time t as a=t^(2)+t+2 where t is in seconds. The particle starts with an initial velocity v-3m/s at t=0. find the velocity of the particle at the end of 5s.

Accleration vs time graph of a particle starting from rest is given. Find the displacement of particle in first 2 seconds.

A particle covers 150m in 8th second starting from rest its acceleration is

The acceleration of a particle starting from rest, varies with time according to the relation A=-aomegasinmegat. . The displacement of this particle at a time t will be

Acceleration of a particle in x-y plane varies with time as a=(2t hati+3t^2 hatj) m//s^2 At time t =0, velocity of particle is 2 m//s along positive x direction and particle starts from origin. Find velocity and coordinates of particle at t=1s.

The acceleration of particle varies with time as shown. (a) Find an expression for velocity in terms of t. (b) Calculate the displacement of the particle in the interval from t = 2 s to t = 4 s. Assume that v = 0 at t = 0.

If a=(3t^2+2t+1) m/s^2 is the expression according to which the acceleration of a particle varies. Then- a=(3t^2+2t+1) (m)/(s^2) Q. The expression for instantaneous velocity at any time t will be (if the particle was initially at rest)-

If acceleration of a particle at any time is given by a = 2t + 5 Calculate the velocity after 5 s, if it starts from rest