Relative Motion of Two Cars: Constant Speed vs Uniform Acceleration

We have two cars starting from the same point at the same initial time:

• Car A moves with constant speed $v_A = 20\,\text{m/s}$.
• Car B starts from rest ($u_B = 0$) and moves with constant acceleration $a_B = 4\,\text{m/s}^2$.

We can describe their positions as functions of time $t$:

1. Position functions

Take the starting point as $x = 0$ at time $t = 0$.

Car A (constant velocity)

For constant speed $v_A$, the position is

$$x_A(t) = v_A t.$$

With $v_A = 20\,\text{m/s}$:

$$x_A(t) = 20 t.$$

Car B (uniform acceleration)

For motion with constant acceleration $a_B$, starting from rest ($u_B = 0$), the position is

$$x_B(t) = u_B t + \tfrac12 a_B t^2 = 0 \cdot t + \tfrac12 a_B t^2 = \tfrac12 a_B t^2.$$

With $a_B = 4\,\text{m/s}^2$:

$$x_B(t) = \tfrac12 (4) t^2 = 2 t^2.$$

So:

• Car A: $x_A(t) = 20 t$
• Car B: $x_B(t) = 2 t^2$

2. When does Car B catch up with Car A?

Car B catches up when they have the same position at the same time:

$$x_A(t) = x_B(t).$$

Substitute the formulas:

$$20 t = 2 t^2.$$

Bring all terms to one side:

$$2 t^2 - 20 t = 0.$$

Factor:

$$2 t (t - 10) = 0.$$

So the solutions are:

$$t = 0 \quad \text{or} \quad t = 10\,\text{s}.$$

• $t = 0$ is the initial moment when both are at the start point.
• The non-trivial solution is:

$$t = 10\,\text{s}.$$

3. Position at the catch-up time

Use either $x_A(t)$ or $x_B(t)$ at $t = 10\,\text{s}$.

Using Car A:

$$x_A(10) = 20 \cdot 10 = 200\,\text{m}.$$

Using Car B:

$$x_B(10) = 2 (10)^2 = 2 \cdot 100 = 200\,\text{m}.$$

They agree, so at $t = 10\,\text{s}$ both cars are 200 m from the starting point.

4. Velocities at the catch-up time

Car A has constant speed $v_A = 20\,\text{m/s}$ at all times.

Car B has acceleration $a_B = 4\,\text{m/s}^2$ starting from rest, so its velocity at time $t$ is

$$v_B(t) = u_B + a_B t = 0 + 4 t = 4 t.$$

At $t = 10\,\text{s}$:

$$v_B(10) = 4 \cdot 10 = 40\,\text{m/s}.$$

So when Car B catches Car A:

• Position: $x = 200\,\text{m}$
• Time: $t = 10\,\text{s}$
• Speeds: $v_A = 20\,\text{m/s}$, $v_B = 40\,\text{m/s}$

5. What the visualization shows

This interactive visualization displays:

• The space–time graph of both cars (position vs. time) in a Cartesian-like plane.
  • Car A: a straight line (constant speed).
  • Car B: an upward-curving parabola (increasing speed under constant acceleration).
• A time cursor moves along the time axis, controlled either by animation or by a time slider.
• At the current time:
  • You see moving dots representing the current positions of Car A and Car B.
  • A vertical guide line shows the current time $t$.
  • The distance between the cars is visualized along the position axis.
• You can adjust:
  • Car A’s speed $v_A$
  • Car B’s acceleration $a_B$
  • The time window displayed, and immediately see how the catch-up time and meeting point change.

Mathematically, the meeting time for general parameters $v_A$ and $a_B$ (with Car B starting from rest at the same point) satisfies

$$v_A t = \tfrac12 a_B t^2 \quad \Rightarrow \quad t = 0 \quad \text{or} \quad t = \frac{2 v_A}{a_B}.$$

The visualization highlights this interaction dynamically.