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.

The displacement x is in centimeter of an oscillating particle varies with time t in seconds as x = 2 cos [0.05pi t+(pi//3)] . Then the magnitude of the maximum acceleration of the particle will be

The displacement x of a particle varies with time 't' as, x=3t^(2)-4t+30. Find the position, velocity and acceleration of the particle at t = 0.

The displacement x of a particle varies with time t as x=4t^(2)-15t+25 Find the position, velocity and acceleration of the particle at t=0. When will the velocity of the particle become zero? Can we call the motion of the particle as one with uniform acceleration ?

The displacement r of a particle varies with time as x=4t^(2)-15t+25 Find the position, velocity and acceleration of the particle at t = 0

The velocity of any particle is related with its displacement As, x = sqrt(v+1) , Calculate acceleration at x = 5m .

The velocity of any particle is related with its displacement As, x = sqrt(v+1) , Calculate acceleration at x = 5m .

Velocity of a particle moving in a straight line varies with its displacement as v= (sqrt(4+4s)) m/s.Find its displacement I the first 2 seconds of its motion.

" If the velocity v of a particle varies as the square of its displacement "x" then the acceleration varies as "