A particle moving on a curve has the position at time `t` given by `x=f'(t) sin t + f''(t) cos t, y= f'(t) cos t - f''(t) sin t` where `f` is a thrice differentiable function. Then prove that the velocity of the particle at time `t` is `f'(t) +f'''(t)`.

Find dy/dx x=e^t (sin t + cos t ),y=e^t(sin t -cos t)

A particle moving along x-axis has acceleration f , at time t , given by f = f_0 (1 - (t)/(T)) , where f_0 and T are constant. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0 , the particle's velocity (v_x) is :

A particle moving along x-axis has acceleration f , at time t , given by f = f_0 (1 - (t)/(T)) , where f_0 and T are constant. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0 , the particle's velocity (v_x) is :

The position of a particle is given by the equation f(t)=t^(3)-6t^(2)+9t where t is measured in second and s in meter. Find the acceleration at time t. What is the acceleration at 4 s?

f is a differentiable function such that x=f(t^2),y=f(t^3) and f'(1)!=0 if ((d^2y)/(dx^2))_(t=1) =

A particle of mass m is at rest the origin at time t= 0 . It is subjected to a force F(t) = F_(0) e^(-bt) in the x - direction. Its speed v(t) is depicted by which of the following curves ?

If x=a(cos t+t sin t) and y=a(sin t - t cos t), then (d^2y)/dx^2

Consider the motion of a particle described by x = a cos t , y= a sin t and z=t . The trajectory traced by the particle as a function of time is

The x and y coordinates of a particle at any time t are given by x = 2t + 4t^2 and y = 5t , where x and y are in metre and t in second. The acceleration of the particle at t = 5 s is

If s=e^(t) (sin t - cos t) is the equation of motion of a moving particle, then acceleration at time t is given by