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=dxdt=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.