The momenta of a body in two perpendicular directions at any time 't' are given by `P_(x)=2t^(2) +6` and `P_(y)=(3t^(2))/(2)+3`. The force acting on the body at `t =2` sec is .

The momentum of a body in two perpendicular direction at any time 't' are given by P_(x) =2t^(2) + 6 and P_(y) = (3t^(2))/2 + 3 . The force acting on the body at t= 2 sec is

The momentum of a body in two perpendicular direction at any time 't' are given by P_(x) =2t^(2) + 6 and P_(y) = (3t^(2))/2 + 3 . The force acting on the body at t= 2 sec is

Three The momentum of a body in two perpendicular direction at any time I are given by P_(x) =2t^(2)+2 +6 and P_(y) =(3t^2)/2+3 The force acting on the body at t = 2 sec is

A body of mass 1kg is moving according to the law x(t)=(5t+4t^(2)+6t^(3))m . The force acting on the body at time t=2s is

The mass of a body is 2.5 Kg. it is in motion and its velocity upsilon after time t is upsilon = (t^3)/(3) +(t^2)/(2) +1 Calculate the force acting on the body at the time t =3 s.

The position of a point in time 't' is given by x = 1 + 2t + 3t^(2) and y = 2 - 3t + 4t^(2) . Then its acceleration at time 't' is

The coordinates of a moving particle at any time t are given by, x = 2t^(3) and y = 3t^(3) . Acceleration of the particle is given by