JEE PYQ: Motion in a Straight Line - Question ID c4d191906f14 (JEE Main 2008)

In this problem we deal with two bodies moving along the same straight line:

• Body 1 starts from rest at $x = 0$ and moves with constant acceleration $a$. Its position as a function of time is given by the kinematic equation for uniformly accelerated motion: $x_1(t) = \tfrac12\,a\,t^2.$
• Body 2 passes through $x = 0$ at $t = 0$ with constant speed $v$. Its position as a function of time is $x_2(t) = v\,t.$

We are asked to plot the difference in their positions, $\Delta x(t) = x_1(t) - x_2(t).$

1. Write the difference explicitly: $\Delta x(t) = x_1(t) - x_2(t) = \tfrac12\,a\,t^2 \;-\; v\,t.$
2. Rearrange as a quadratic in $t$: $\Delta x(t) = \tfrac12\,a\,t^2 \;-\; v\,t = \tfrac12\,a\,\bigl(t^2 - \tfrac{2v}{a}\,t\bigr).$
3. Complete the square to see the shape: $t^2 - \tfrac{2v}{a}\,t = \Bigl(t - \tfrac{v}{a}\Bigr)^2 \;-\; \Bigl(\tfrac{v}{a}\Bigr)^2.$ Hence $\Delta x(t) = \tfrac12\,a\Bigl[\Bigl(t - \tfrac{v}{a}\Bigr)^2 - \Bigl(\tfrac{v}{a}\Bigr)^2\Bigr] = \tfrac12\,a\,\Bigl(t - \tfrac{v}{a}\Bigr)^2 \;-\; \tfrac{v^2}{2a}.$ This is a parabola opening upwards, with its vertex at $t = \tfrac{v}{a},\quad \Delta x = -\tfrac{v^2}{2a}.$
4. Interpret the graph:
   • At $t=0$, $\Delta x(0)=0$.
   • For $0 < t < \tfrac{v}{a}$, $\Delta x(t)$ decreases (becomes negative) because the second body is moving faster initially.
   • At $t = \tfrac{v}{a}$, $\Delta x$ reaches its minimum value $-\tfrac{v^2}{2a}$.
   • For $t > \tfrac{v}{a}$, $\Delta x(t)$ increases without bound as the accelerated body overtakes the constant-speed one.
5. Compare with the given options: Only option B shows a parabola that starts at zero, dips to a negative minimum, and then rises again.

Many students mistakenly think the difference $x_1 - x_2$ should be a straight line or a parabola that never goes negative. Remember:

• The initially faster body (constant speed) pulls ahead, making $\Delta x$ negative at first.
• Only after the accelerated body “catches up” does $\Delta x$ become positive again.
• Always write the explicit quadratic form and check its vertex to avoid choosing the wrong graph.