Ncert Exercises & Solutions

A graph of x versus t is shown in the figure.

(a) The particle was released from rest at t = 0.
(b) At B, the acceleration a > 0.
(c) Average velocity for the motion between A and D is positive.
(d) The speed at D exceeds that at E.

Choose the correct alternatives:

1. (b, d)

2. (a, b)

3. (b, c)

4. (a, d)

(4) Hint: The slope of the d-t graph gives the velocity.

Step 1: Find the slope of the graph at the concerning points.

As per the diagram, at point A the graph is parallel to the time axis hence, $v = \frac{dx}{dt} = 0$ . As the starting point is A, hence, we can say that the particle is starting from the rest. From the graph, it is clear that

|slope at D| > |slope at E|

Hence, the speed at D will be more than at E.

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