A stone is falling from a mountain. The velocity of the stone is given by V = 9.8t. Draw its graph and find the velocity of the stone ‘4’ seconds after start.

A particle starts from the origin and moves along a straight line. If the velocity of the particle at time t seconds be 10t cm/s, find the distance traversed during first 4 seconds after the start.

A balloon is rising upwards with an acceleration. A stone is dropped frome the balloon, is at a height of 50.4 m. The stone reaches the ground after 6s. What was the velocity of the balloon when the stone was dropped?

The velocity of a moving particle is given by v = 6 +18t + 9t^(2) (x in metre t in second ) what is its acceleration at t = 2 s?

A particle starting from a point A and moving with a positive constant acceleration along a straight line reaches another point B is time T. Suppose that the initial velocity of the particle is u gt0 and P is the midpoint of the line AB. If the velocity of the particle at point P is v_(1) and if the velocity at time (T)/(2) is v_(2) , then -

From the top of a tower, 80 m high from the ground, a stone is thrown in the horizontal direction with a velocity of 8ms^(-1) . The stone reaches the ground after a time 't' and falls at a distance of 'd' from the foot of the tower. Assuming g=10ms^(-2) , the time t and distance d are given respectively by

A piece of stone was dropped from a stationary baloon. The stone covered 13.9 m during the last (1)/(7) s of its descent. Find the height of the balloon and the velocity of the stone whe it strikes the ground. [g = 9.8 "m.s" ^(-2) ]

If the acceleration of a particle at time t seconds be 1.2t^(2)ft//s^(2) and velocity of the particle at time 2 seconds be 2.2 ft/s, then find its velocity at time 5 seconds.