For the body starting from rest

$${x_1} = 0 + {1 \over 2}a{t^2} \Rightarrow {x_1} = {1 \over 2}a{t^2}$$

For the body moving with constant speed

$${x_2} = vt$$

$$\therefore$$ $${x_1} - {x_2} = {1 \over 2}a{t^2} - vt$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{d\left( {{x_1} - {x_2}} \right)} \over {dt}} = at - v$$

at $$t=0,$$ $$\,\,\,\,\,\,{x_1} - {x_2} = 0$$, so graph should start from origin.

For $$at < v;$$ the slope is negative that means $${x_1} - {x_2}$$ < 0 so initially velocity of 1st body is less than second body and velocity of 1st body is increasing gradually.

For $$at = v;$$ the slope is zero. So $${x_1} - {x_2}$$ = 0 it means here velocity of both the bodies are same.

For $$at > v;$$ the slope is positive. So $${x_1} - {x_2}$$ > 0 it means here velocity of first body is greater than second body.

We know the relation between distance and time is.

$$S = ut + {1 \over 2}a{t^2}$$, which is a equation parabola. So the graph should be a parabola.

These characteristics are represented by graph $$(b).$$