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, . 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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