A particle moves so that x=2+27t-`t^(3)`. The direction of motion reverses after moving a distance of . . . Units.

A particle moves so that s=6+48t- t^3 . The direction of motion reverses after moving a distance of

A particle moves so that s=6+48t- t^3 . The direction of motion reverses after moving a distance of:- (a)63 (b)104 (c)134 (d)288

The distance form a fixed point O of a particle pmoving in a straight line from O is given by s=16+48t-t^3 . The direction of motion of the particle after t=4 sec is

The position vector vec(r) of a moving particle at time t after the start of the motion is given by vec(r) = (5+20t)hat(i) + (95 + 10t - 5 t^(2)) hat(j) . At the t = T, the particle is moving at right angles to its initial direction of motion. Find the value of T and the distance of the particle from its initial position at this time.

A particle moves according to the law s=t^(3)-9t^(2)+24t . The distance covered by the particle before it first comes to rest is -

A particle is travelling along a straight line OX. The distance x (in metres) of the particle from O at a time t is given by x=37+27t-t^(3) where t is time in seconds. The distance of the particle from O when it comes to rest is

A particle moves in a straight line, so that after t second, the distance x from a fixed point O on the line is given by x=(l-2)^(2)(t-5) . Then