Acceleration and Retardation

A car moves in a straight line, the car accelerates from rest with a constant acceleraation alpha on a straight foad. After gaining a velocity v , the car moves with that velocity for somerime. Then car decelerates with a retardation beta , If the total distance covered by the car is equal to s find the total time of its motion.

A lift performs the first part of its ascent with uniform acceleration a and the remaining with uniform retardation 2a. If t is the time of ascent, find the depth of the shaft.

A train stopping at two stations 4 km apart takes 4 min on the journey from one of the station to the other. Assuming that it first accelerates with a uniform acceleration x and then that of uniform retardation y, prove that 1/x + 1/y = 2.

Assertion : A sphere is placed in pure rolling condition over a rough inclined surface.Then, force of friction will act in downward direction Reason : Angular acceleration (actually retardation) due to friction is anti-clockwise.

A block of mass m is attached with a massless in string. Breaking strength of string is 4mg. Block is moving up. The maximum acceleration and maximum retardation of the block can be

Acceleration of a particle has a value 'a' for a time t. It is followed immediately by a retardation of magnitude 'a' for time t/2. Consider this as one cycle. Initial velocity of particle was zero. The displacement of the particle after n such cycles in succession is:

A body moves for a total of nine second starting from rest with uniform acceleration and then with uniform retardation, which is twice the value of acceleration and then stop. The duration of uniform acceleration is

If retardation produced by air resistances to projectile is one-tenth of acceleration due to gravity, the time to reach maximum height approximately-

A body falling from a vertical height of 10 m pierces through a distance of 1 m in sand. It faces and average retardation in sand. It faces an average retardation in sand equal to (g = acceleration due to gravity)

The friction of the air causes a vertical retardation equal to 10% of the acceleration due to gravity (take g=10ms^(-2)) The maximum height will be decreased by: