A graph of x versus t is shown in Fig. 3.3. Choose correct alternatives from below.
(a) The particle was released from rest at t = 0.
(b) At B, the acceleration a > 0.
(c) At C, the velocity and the acceleration vanish.
(d) Average velocity for the motion between A and D is positive.
(e) The speed at D exceeds that at E.
The correct answer is the option:
(a) The particle was released from rest at t = 0.
(c) At C, the velocity and acceleration vanish.
(e) The speed at D exceeds that at E.
Explanation: Now, we know that,
Slope of the x-t graph gives us
Verification of opt(a)-
is zero or particle is at rest at A since graph (x-t) is parallel to the time axis.
Slope increases after A and hence velocity also increases.
Verifying option (c) and rejecting opt (b)-
Now, or v = 0 since the tangent at B & C is graph (x-t), viz., parallel to the time axis. Hence, acceleration = 0.
Verifying opt (e)-
Speed at D is greater than speed at E since the slope at D is greater at D than at E.
Rejecting opt (d)-
Average velocity at A is zero as graph (x-t) is parallel to time axis, also displacement is negative at D, which makes it clear that the velocity at D is also negative.