Write the position equations

• First body: starts from rest at $x=0$ with constant acceleration $a$ in the positive $x$-direction. $x_1(t)=\frac{1}{2}at^2$

• Second body: passes through $x=0$ at $t=0$ with constant speed $v$ in the positive $x$-direction. $x_2(t)=vt$

Find the required quantity

We need the graph of $x_1-x_2=\frac{1}{2}at^2-vt$

Let $y(t)=x_1-x_2=\frac{1}{2}at^2-vt$

Analyze the nature of the graph

This is a quadratic function of $t$ with positive coefficient of $t^2$, so it is an upward opening parabola.

Check important points

• At $t=0$, $y(0)=0$ So the graph passes through the origin.

• Initial slope: $\frac{dy}{dt}=at-v$ At $t=0$, $\left.\frac{dy}{dt}\right|_{t=0}=-v<0$ So just after $t=0$, $x_1-x_2$ becomes negative.

• Minimum occurs when $\frac{dy}{dt}=0 \Rightarrow at-v=0 \Rightarrow t=\frac{v}{a}$ Thus the curve first decreases, reaches a minimum, then increases.

• It again becomes zero when $\frac{1}{2}at^2-vt=0$ $t\left(\frac{1}{2}at-v\right)=0$ So, $t=0 \quad \text{or} \quad t=\frac{2v}{a}$

Hence the graph:

• starts from origin,
• initially goes below the time axis,
• reaches a minimum,
• rises and crosses the axis again,
• then becomes positive.

Match with the options

Therefore the correct graph is the one representing an upward opening parabola passing through the origin, initially decreasing into negative values, and later crossing back to positive values.

That corresponds to Option B.