The distance (s) travelled by a body in time (t) is given by `s = ut + (1)/(2)at^(2)`.
How is the formula modified when
(i) body starts from rest , (ii) motion of body is uniform.

If the distance s travelled by a body in time t is given by s = a/t + bt^2 then the acceleration equals

The equation of distance travelled by a body s with time t is given as S=A(t+B)+Ct^(2) . The dimension of B is given by

The distance travelled by a body, falling freely from rest in t=1s, t=2s and t=3s are in the ratio of

Establish the eqaution for position time relation (i.e., s = ut + (1)/(2)at^(2) ) using velocity time graph. How is the equation modified when (i) a body is just dropped from some height, (ii) a body is thrown vertically upwards with some velocity ?

When body is at rest or it is in uniform motion, no force act on it.

The distance (S) travelled varies with time (t) for four different bodies as given below. In which case is the acceleration of the body minimum?

A body slides down on an inclined plane of inclination 37^(@) with horizontal. The distance travelled by the body in time t is given by s = t^(2) . Find the friction coefficient between the body and the incline.

The displacement of body is given by s = 4 + 2t^4 . The acceleration of the body at the end of 1 s from the start is :

(a) Draw distance time graph when a body is at rest. How is the graph modified when the body is moving with a uniform velocity ? (b) A car moving at 36 km//h is brought to rest in 0.1 km . What is the retardation ?

Acceleration of a body is given by a = 2sqrt(t) , where t is the time. If the body starts from rest what will be its velocity and acceleration 4 s later ? [ SI units are to be used ]