For a body moving withuniform acceleration 'a', initial and final velocities in a time interval are 'u' and respectively.Then,its average velocity in the time interval T is

For body moving with uniform acceleration a , initial and final velocities in a time interval t are u and v respectively. Then its average velocity in the time interval t is

A particle moves in a straight line with uniform acceleration. Its velocity at time t=0 is v_1 and at time t=t is v_2. The average velocity of the particle in this time interval is (v_1 + v_2)/2 . Is this statement true or false?

Show that area under the velocity-time graph of an object moving with constant acceleration in a straight line in certain time interval is equal to the distance covered by the object in the interval.

Statement-I : A particle moves in a straight line with constant acceleration. The average velocity of this particle cannot be zero in any time interval. Statement-II : For a particle moving in straight line with constant acceleration, the average velocity in a time interval is (u+v)/(2) , where u and v are initial and final velocity of the particle of the given time interval.

A particle move with a velocity v=alpha t^(3) along a straight line.The average velocity in time interval t=0 to t=T is

A particle is moving on a straight line and its average velocity is found to be zero in an interval of time.

If magnitude of average speed and average velocity over a time interval are same, then

Show that the area under the velocity-time graph of an object moving with constant acceleration in a straight line in certain time interval is equal to the distance covered by the object in that interval.

An object is moving with uniform acceleration a. Its initial velocity is u and after time t its velocity is v. The equation of its motion is v = u + at. The velocity (along y-axis) time (along x-axis) graph shall be a straight line