Motion of particle is given by equation
`s=(5t^(3)+3t^(2)+8t+17)m`
The value of acceleration of the particle at t = 2 s is

Motion of a particle is given by equation S = (3t^(3) + 7 t^(2) + 14 + t + 8) m, The value of acceleration of the particle at t = 1 sec, is:

A particle moves along a straight line such that its displacement at any time t is given by s = 3t^(3)+7t^(2)+14t + 5 . The acceleration of the particle at t = 1s is

The velocity time relation of a particle is given by v = (3t^(2) -2t-1) m//s Calculate the position and acceleration of the particle when velocity of the particle is zero . Given the initial position of the particle is 5m .

If the velocity of a paraticle moving along x-axis is given as v=(4t^(2)+3t=1)m//s then acceleration of the particle at t=1sec is :

Velocity of a particle is given as v = (2t^(2) - 3)m//s . The acceleration of particle at t = 3s will be :

the angular velocity omega of a particle varies with time t as omega = 5t^2 + 25 rad/s . the angular acceleration of the particle at t=1 s is

The displacement of a moving particle is given by, x=at^3 + bt^2 +ct + d . The acceleration of particle at t=3 s would be

The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 2/3s is

Acceleration of a particle is given by a= 3t^(2)+2t +1 Where t is time . Find (i) its velocity at time t assuming velocity to be 10 m/s at t = 0. (ii) its position at time t assuming that the particle is at origin at t = 0. (iii) What is the average speed of the particle in the duration t = 0 to t = 1s? (iv) What is the average acceleration of the particle in the duration t = 0 to t = 1s?

The velocity of a particle is given by v=12+3(t+7t^2) . What is the acceleration of the particle?