Kinematics Test - 15

Average velocity: It is defined as the ratio of displacement to time taken by the body.

According to this problem, we need to identify the graph which is having the same displacement for two timings. When there are two timings for same displacement, the corresponding velocities should be in opposite directions. As shown in graph (b), the first slope is decreasing that means particle is going in one direction and its velocity decreases, becomes zero at highest point of the curve and then increasing in backward direction. Hence the particle

return to its initial position. So, for one value of displacement there are two different points of time and we know that slope of x, x-t graph gives us the average velocity. Hence, for one time, slope is positive then average velocity is A also positive and for other time slope is negative then average velocity is also negative. As there are opposite velocities in the interval 0 to T, hence average velocity can only vanish in option (B)

This can be seen in the figure given alongside.

As shown in the graph,  OA = BT (same displacement) for two different points of time

From graph, whent = 0, the particle is released from rest at A, hence, v = 0.

At B, the graph is parallel to time axis, hence velocity is constant there. The acceleration a is zero.

At C, the graph changes slope, where velocity and acceleration vanish.

Average velocity for motion between A and D is negative, because the value of x is decreasing with time t. The slope of graph (which represents speed) is more at D than at E.