A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant a

Q: Why does the graph dip below the t-axis initially?

Because the second body starts with a non-zero velocity while the first body starts from rest. Initially, the second body covers more distance, making $(x_1 - x_2)$ negative.

Q: What does the lowest point on the graph represent?

It represents the moment of maximum separation between the two bodies, which occurs when their velocities are exactly equal.

Q: Why is the graph a parabola?

Because the relative position equation is a quadratic function of time ($\frac{1}{2}at^2 - vt$), and all quadratic functions plot as parabolas.

Q: Will the first body always overtake the second body?

Yes, as long as it has a constant positive acceleration, its velocity will eventually exceed the constant velocity of the second body, allowing it to catch up and overtake.

Q: What does the second x-intercept mean?

It is the exact time when the first body catches up to the second body, making their relative position zero again.

Question

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?

Options:

(A)

(B)

(C)

(D)

Sigma Percentile · Accepted Answer

The beauty of kinematics lies in how perfectly physical motion translates into mathematical geometry. In this problem, we are witnessing a classic race between two fundamental types of motion: constant speed and constant acceleration.

Imagine you are standing at the starting line of a drag race. Car 1 starts from rest but has a powerful engine providing constant acceleration. Car 2, on the other hand, doesn't start from rest; it comes flying past the starting line at a constant speed exactly when the clock starts.

Our goal is to visualize the distance between them as the race unfolds. Let's break down their journeys mathematically.

### Analyzing the Setup

First, we need to write down the position of each body as a function of time. For the first body, it starts from rest ($u = 0$) at the origin ($x = 0$) and moves with a constant acceleration $a$. Using the second equation of motion, its position $x_1(t)$ is given by:

$$x_1(t) = \frac{1}{2}at^2$$

The second body passes the origin at $t = 0$ with a constant speed $v$. Since there is no acceleration for this body, its position $x_2(t)$ is simply:

$$x_2(t) = vt$$

Now, the question asks us to find the graph of $(x_1 - x_2)$ as a function of time. Let's define this relative position as $\Delta x(t)$.

### The Master Equation

By subtracting the two position functions, we get our master equation for the relative distance:

$$\Delta x(t) = \frac{1}{2}at^2 - vt$$

Take a close look at this equation. It is a quadratic function of time $t$. In the world of coordinate geometry, every quadratic function represents a **parabola**. Because the coefficient of $t^2$ is positive ($\frac{1}{2}a > 0$), we know this parabola must open upwards.

But where does it start, and where does it go? Let's find out by checking its initial state.

### Tracing the Parabola

At the exact moment the race begins ($t = 0$), we can substitute $t = 0$ into our equation:

$$\Delta x(0) = \frac{1}{2}a(0)^2 - v(0) = 0$$

This tells us the graph must start exactly at the origin $(0,0)$. Both bodies are at the same position initially.

Next, let's look at the initial slope of the graph. The slope represents the relative velocity, which is the derivative of our function:

$$v_{	ext{rel}} = \frac{d(\Delta x)}{dt} = at - v$$

At $t = 0$, the initial relative velocity is $-v$. This is a crucial insight! It means the slope is negative right at the start. Physically, this makes perfect sense: Car 2 is already moving at speed $v$, while Car 1 is just starting from zero. Car 2 immediately pulls ahead, making the relative position $(x_1 - x_2)$ negative.

Therefore, our graph must dip below the horizontal axis initially.

### The Turning Point and Overtake

As Car 1 continues to accelerate, it will eventually reach the same speed as Car 2. This happens when $at - v = 0$, or $t = \frac{v}{a}$. At this exact moment, the distance between them is at its maximum, and our parabola hits its lowest point (the vertex).

After this point, Car 1 is moving faster than Car 2 and starts closing the gap. Eventually, Car 1 will catch up and overtake Car 2. We can find this overtaking time by setting $\Delta x(t) = 0$:

$$\frac{1}{2}at^2 - vt = 0$$

Factoring out $t$, we get $t = 0$ (the start) and $t = \frac{2v}{a}$ (the overtake). At $t = \frac{2v}{a}$, the parabola crosses the horizontal axis again and shoots upwards into the positive region as Car 1 leaves Car 2 behind.

**Conclusion:** Our mathematical journey reveals an upward-opening parabola that starts at the origin, dips into the negative region, reaches a minimum, and then rises to cross the axis again. This perfectly matches the curve shown in **Option (b)**.

Select Answer:

Visualized Solution

$x_{1} (t)$ and $x_{2} (t)$

$Δ x (t)$

At $t = 0$

Initial Slope

Minimum Point

Overtaking

Conclusion

The Sigma Insight: Motion Graphs

Analyzing the Setup

The Master Equation

Tracing the Parabola

The Turning Point and Overtake

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