A particle crosses points A, B, C in a straight line at a constant acceleration. The particle travels from A to B in time `t_(1)` and B to C in time `t_(2)` . If AB = a , BC = b then prove that the acceleration of the particle is `2(bt_(1)-at_(2))//t_(1)t_(2)(t_(1)+t_(2))`.

O is fixed point on the straight line traced out by a moving particle. If the distance of the particle from O at time t be (a cos nt + b sin nt ) [ a,b and n are constants]., show that the acceleration of the particle varies as its distance from O.

A particle is moving in a straight line. At time, the distance between the particle from its starting point is given by x=t-6t^(2)+t^(3) . lts acceleration will be zero at

A particle of mass m accelerating uniformly has velocity v at time t_1 . What is work done in time t?

If the distance from the initial point at a time t is given by x = t - 6t^2 + t^3 : then at what time the acceleration will be zero?

The displacement x of a particle in rectilinear motion at time t is represented as x^(2) = at^(2) +2bt+c , where a ,b, and c are constants. Show that the acceleration of this particle varies (1)/(x^(3)) .

The x and y coordiates of a particle at any time t are given by x = 7t + 14 t^(2) and y = 5t, where x and y are in metre and t is in second. The acceleration of the particle at t = 5 s is

A particle starting from a point A and moving with a positive constant acceleration along a straight line reaches another point B is time T. Suppose that the initial velocity of the particle is u gt0 and P is the midpoint of the line AB. If the velocity of the particle at point P is v_(1) and if the velocity at time (T)/(2) is v_(2) , then -