A body is at rest at $$x = 0$$. At $$t = 0$$, it starts moving in the positive $$x$$-direction with a constant acceleration. At the same instant another body passes through $$x = 0$$ moving in the positive $$x$$ direction with a constant speed. The position of the first body is given by $$x_1(t)$$ after time '$$t$$' and that of the second body by $$x_2(t)$$ after the same time interval. Which of the following graphs correctly describes $$(x_1 - x_2)$$ as a function of time '$$t$$'?

The first body starts from rest with uniform acceleration $$a$$, so after time $$t$$ its displacement from the origin is

$$x_1(t)=\tfrac12 a t^{2}\qquad -(1)$$

The second body crosses the origin at $$t=0$$ with a constant speed $$v$$, hence

$$x_2(t)=v\,t\qquad -(2)$$

The required function is the difference of the two displacements:

$$x_1-x_2=\tfrac12 a t^{2}-v t=t\bigl(\tfrac12 a t-v\bigr)\qquad -(3)$$

Equation (3) is a quadratic in $$t$$ whose leading coefficient $$\tfrac12 a$$ is positive, so its graph is an upward-opening parabola.

Key features of the parabola:

• At $$t=0$$, $$x_1-x_2=0$$ (the two bodies are together).
• The initial slope is $$\left.\frac{d}{dt}(x_1-x_2)\right|_{t=0}= -v$$, i.e. the curve leaves the origin with a negative slope.
• The vertex occurs when $$\dfrac{d}{dt}(x_1-x_2)=a t-v=0\;\Longrightarrow\;t=\dfrac{v}{a}$$. The minimum value there is $$-\dfrac{v^{2}}{2a}$$, so the curve lies below the $$t$$-axis between $$t=0$$ and $$t=\dfrac{2v}{a}$$.
• It crosses the $$t$$-axis again when $$x_1-x_2=0\Rightarrow t=0$$ or $$t=\dfrac{2v}{a}$$ and thereafter remains positive.

Therefore the correct plot begins at the origin, dips below the $$t$$-axis, attains a minimum at $$t=\dfrac{v}{a}$$, emerges to cut the axis at $$t=\dfrac{2v}{a}$$, and then rises indefinitely — exactly the shape shown in Option B.

Option C which is: the upward-opening parabola starting at the origin with an initial negative slope and crossing the axis once more at $$t=\dfrac{2v}{a}$$