If acceleration (a)=2t+1, where t is time.Find displacement as function of time,Given that initally particle is at rest

A particle is moving under constant acceleration a=k t . The motion starts from rest. The velocity and displacement as a function of time t is

A starts from rest, with uniform acceleration a. The acceleration of the body as function of time t is given by the equation a = pt, where p is a constant, then the displacement of the particle in the time interval t = 0 to t=t_(1) will be

A particle starts from rest with a time varying acceleration a=(2t-4) . Here t is in second and a in m//s^(2) The velocity time graph of the particle is

A particle starts from rest and moves with acceleration a which varies with time t as a=kt where k is a costant. The displacement s of the particle at time t is

It is given that t= px^(2) + qx , where x is displacement and t is time. The acceleration of particle at origin is

A particle moves in the x-y plane according to the law x=t^(2) , y = 2t. Find: (a) velocity and acceleration of the particle as a function of time, (b) the speed and rate of change of speed of the particle as a function of time, (c) the distance travelled by the particle as a function of time. (d) the radius of curvature of the particle as a function of time.

A point moves such that its displacement as a function of times is given by x^(2)=t^(2)+1 . Its acceleration at time t is

A particle starts from rest, moving in a straight line with a time varying acceleration ( "in"m//s^(2) ): a= 30 -(15)/(2) sqrt(t) , where time is in seconds . The displacement of the particle in 4 seconds is .

A body starts from rest & moves with constant acceleration a_1 for time t_1 then it retards uniformly with a_2 , in time t_2 and finally comes at rest. Find (t_1)/(t_2)