If velocity of a particle is `v=bs^(2)` ,where `b` is constant and `s` is displacement. The acceleration of the particle as function of s is?

The displacement of a particle is s = (a+bt)^(6) , where a and b are constants. Find acceleration of the particle as a function of time.

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 "

A particle of unit mass is moving along x-axis. The velocity of particle varies with position x as v(x). =alphax^-beta (where alpha and beta are positive constants and x>0 ). The acceleration of the particle as a function of x is given as

For a particle moving along a straight line, its velocity 'v' and displacement 's' are related as v^(2) = cs , here c is a constant. If the displacement of the particle at t = 0 is zero, its velocity after 2 sec is:

. If the velocity of a particle is (10 + 2 t 2) m/s , then the average acceleration of the particle between 2s and 5s is

If the velocity of a particle is (10+2t^(2)m/s ,then the average acceleration of the particle between 2s and 5s is (in m/s^2)

The displacement of a particle after time t is given by x = (k // b^(2)) (1 - e^(-bi)) . Where b is a constant. What is the acceleration fo the particle ?