A body starts with a velocity `(2hati+3hatj+11hatk)m//s` and moves with an acceleration `(5hati+5hatj-5hatk)m//s^(2)`. What is its velocity after `0.2 sec`?

The velocity of an object at t =0 is vecv_(0) =- 4 hatj m/s . It moves in plane with constant acceleration veca = ( 3hati + 8 hatj) m//s^(2) . What is its velocity after 1 s?

A particle P is at the origin starts with velocity u=(2hati-4hatj)m//s with constant acceleration (3hati+5hatj)m//s^(2) . After travelling for 2s its distance from the origin is

A body of mass 3 kg moving with a velocity (2hati+3hatj+3hatk) m/s collides with another body of mass 4 kg moving with a velocity (3hati+2hatj-3hatk) m/s. The two bodies stick together after collision. The velocity of the composite body is

A mosquito is moving with a velocity vecv = 0.52t^(2)hati + 3t hatj+ 9hatk m//s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s?

A force of (5hati+6hatj) N makes a body to move on a rough surface with a velocity of (4hatj-2hatk) m/s. What is the power?

A particle of mass 2kg is moving in free space with velocity vecv_0=(2hati-3hatj+hatk)m//s is acted upon by force vecF=(2hati+hatj-2hatk)N . Find velocity vector of the particle 3s after the force starts acting.

If angular velocity of a point object is vecomega=(hati+hat2j-hatk) rad/s and its position vector vecr=(hati+hatj-5hatk) m then linear velocity of the object will be

A constant force of (2hati+3hatj+5hatk) N produces a displacement of (3hati+2hatj+2hatk) m. Then work done is

A particle moving with velocity (2hati-3hatj) m/s collides with a surface at rest in xz–plane as shown in figure and moves with velocity (2hati+2hatj) m/s after collision. Then coefficient of restitution is :