A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate of 1 m/`s^(2)`, at the same time the man starts running with uniform velocity of 10 m/s. What is the minimum time in which the man catches the bus? (8s)`(AS_(7))`

A bus starts from rest with an acceleration of 1 m//s^(2) A man who is 48 m behind the bus starts with a uniform velocity of 10 m s^(-1) . Then after how much time the man will catch the bus ?

A student is standing at a distance of 50 metre from the bus. As soon as the bus begins its motion with an acceleration of 1m//s^(2) , the student starts running towards the bus with a uniform velocity u. assuming the motion to be along a straight road, the minimum value of u. so that the student is able to catch the bus is

A man runs towards a mirror at a rate of 6 m/s. If we assume the mirror to be at rest, his image will have a velocity

A racing car has a uniform acceleration of 4 m//s^(2) . What distance will it cover in 10s after start ?

A man runs at a speed of 4 ms to overtake a standing bus. When he is 6m behind the door (a ) the bus moves forward and continues moving with constant acceleration of 1.2 m/s^2 . The time after which the man is able to catch the bus is :

A man weighing 60 kg is a lift moving down with an acceleration of 1.8 m s^(-2) . The force exerted by the floor on him is

The acceleration a' in ms-2 of a particle is given by a = 3t^2 +2t+2, where is the time. If the particle starts with a velocity v = 2 m s-at i=0 then velocity at the end of 2 sec is:

Two bodies begin to move from rest in the same straight line at the same instant of time from the same point. The first moves with constant velocity of 40 ms and the second starts with uniform acceleration of 4 ms. The time which the distance between them is maximum is :

A man is at a distance from the bus: when the bus begins to move with a constant acceleration 'a', with what minimum velocity the man should run towards the bus so as to catch it?