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

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

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

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 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.

If the velocity of a body related to displacement x is given by v=sqrt(5000+24x) m/s, then the acceleration of the body is _______ m//s^(2) .

The velocity v of moving object varies with displacement u as v=3x^(2)-2 ,then the acceleration at x=2m is

If v = sqrt(5000+24x) m/s then calculate acceleration.

Velocity of a particle moving in a straight line varies with its displacement as v = sqrt(4+4x)m//s where x is displacement. Displacement of particle at time t = 0 is x = 0. Find displacement (in m) of particle in 2 sec.

The velocity of particle is given by v=sqrt(gx) . Find its acceleration.

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 .