Displacement `s` of a particle at time `t` is expressed as `s=1/2t^3-6t`. Find the acceleration at the time when the velocity vanishes (i.e., velocity tends to zero).

If S=t^3+6t^2-36t+100 . Determine acceleration and displacement at the time when velocity vanishes

The displacement (s ) of a particle depends on time (t) as s = 2at^(2)-bt^(3) . Then

The velocity of a particle moving along the positive direction of x-axis is v= Asqrt(x) , where A is a positive constant. At time t = 0 , the displacement of the particle is taken as x = 0 . (i) Express velocity and acceleration as functions of time , and (ii) find out the average velocity of the particle in the period when it travels through a distance s ?

If the displacement of a particle changes with time as sqrt(x)=t+3 , then the velocity of the particle will be proportional to

If the acceleration of a particle at time t seconds be 1.2t^(2)ft//s^(2) and velocity of the particle at time 2 seconds be 2.2 ft/s, then find its velocity at time 5 seconds.

The displacement of a particle starting from rest (at t = 0) is given by s = 6t^(2)-t^(3) . The time in second at which the particle obtain zero velocity again is

Displacement x and time t, in a rectiIinear motion of a partical are reIated as t = sqrt(x)+3 . Here x is measured in metre and t in second. Find the displacement of the partical when its velocity is zero.

A particle moves in a straight line such that its distance x from a fixed point on it at any time t is given by x=(1)/(4)t^(4)-2t^(3)+4t^(2)-7 Find the time when its velocity is maximum and the time when its acceleration is minimum .