Get Answers to all your Questions

header-bg qa

A graph of x versus t is shown in Fig. 3.3. Choose the 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) The average velocity for the motion between A and D is positive.
(e) The speed at D exceeds that at E.

Answers (1)

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,

The slope of the x-t graph gives us V=\frac{dx}{dt}

Verification of opt(a)-

\frac{dx}{dt} Is zero, or particle is at rest at A since graph (x-t) is parallel to the time axis?

Slope \frac{dx}{dt} increases after A and hence velocity also increases.

Verifying option (c) and rejecting opt (b)-

Now, \frac{dx}{dt} 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)-

The speed at D is greater than the speed at E since the slope at D is greater at D than at E.

Rejecting opt (d)-

The average velocity at A is zero as graph (x-t) is parallel to the time axis, and displacement is negative at D, which makes it clear that the velocity at D is also negative.

Posted by

infoexpert22

View full answer