For a body in `SHM` the velocity is given by the relation `V = sqrt(144-16x^(2))ms^(-1)`. The maximum acceleration is

If the velocity of a particle is given by v=(180-16x)^((1)/(2))(m)/(s) , then its acceleration will be

The velocity of a particle is given by v=(2t^(2)-3t+10)ms^(-1) . Find the instantaneous acceleration at t = 5 s.

For a particle in SHM the amplitude and maximum velocity are A and V respectively. Then its maximum acceleration is

The velocity of a particle varies with its displacement as v=(sqrt(9-x^2))ms^-1 . Find the magnitude of the maximum acceleration of the particle.

A particle is executing a linear S.H.M. Its velocity at a distance x from the mean position is given by v^2=144-9x^2 . The maximum velocity of the particle is

A body executes SHM , such that its velocity at the mean position is 1ms^(-1) and acceleration at exterme position is 1.57 ms^(-2) . Calculate the amplitude and the time period of oscillation.

A body of mass 1//4 kg is in SHM and its displacement is given by the relation x = 0.05 Sin(20t +(pi)/(2))m . If is in seconds, the maximum force acting on the particle is

A body is projected from the ground with a velocity v = (3hati +10 hatj)ms^(-1) . The maximum height attained and the range of the body respectively are (given g = 10 ms^(-2))

The maximum velocity of a particle executing SHM is 100cms^(-1) and the maximum acceleration is 157cms^(-2) determine the periodic time

The rod of length l=1 m rotates with an angular velocity omega=2 rads^(-1) an the point P moves with velocity v=1ms^(-1) and acceleration a=1ms^(-2) . Find the velocity and acceleration of Q .