The velocity (v) – time (t) plot of the motion of a body is shown below:

The acceleration (a) – time (t) graph that best suits this motion is :

The correct option is Option 1 (represented by the first graph).

Step-by-Step Explanation:

To determine the corresponding acceleration–time ($a$–$t$) graph from the given velocity–time ($v$–$t$) graph, we use the fundamental definition of acceleration.

Acceleration is the rate of change of velocity with respect to time. Mathematically, it is defined as the derivative of velocity:
$a=\frac{dv}{dt}$
This derivative corresponds to the slope of the tangent to the curve on a velocity–time ($v$–$t$) graph at any given instant.

Let us analyze the motion by breaking the velocity–time graph into five distinct consecutive time intervals:

1. First Interval (Initial horizontal line on the time axis):
The velocity is constant and remains at zero ($v=0$). Since the velocity is not changing, the slope of the line is zero.
Therefore, the acceleration is:
$a=0$
This is represented by a horizontal segment along the time-axis in the $a$–$t$ graph.

2. Second Interval (Upward sloping straight line):
The velocity increases linearly with time. For a straight line sloping upwards, the slope is constant and positive.
Therefore, the acceleration is constant and positive:
$a>0$
This is represented by a flat, positive rectangular step above the time-axis in the $a$–$t$ graph.

3. Third Interval (Horizontal flat peak line):
The velocity remains constant at its maximum value. For a horizontal straight line, the slope is zero.
Therefore, the acceleration is:
$a=0$
This is represented by a horizontal segment along the time-axis in the $a$–$t$ graph.

4. Fourth Interval (Downward sloping straight line):
The velocity decreases linearly with time. For a straight line sloping downwards, the slope is constant and negative.
Therefore, the acceleration is constant and negative:
$a<0$
This is represented by a flat, negative rectangular step below the time-axis in the $a$–$t$ graph.

5. Fifth Interval (Final horizontal line on the time axis):
The velocity is constant and remains at zero. The slope of the line is zero.
Therefore, the acceleration is:
$a=0$
This is represented by a final horizontal segment along the time-axis in the $a$–$t$ graph.

Conclusion:
Combining these intervals, the acceleration–time graph starts at zero, jumps to a constant positive value, returns to zero, drops to a constant negative value, and finally returns to zero. This sequence corresponds exactly to the rectangular pulses shown in Option 1.