Animated Solution for Physics - Kinematics: A material particle is chasing another one and both of them are moving on the same straight line. After they pass a particular point, their velocities $v$ vary with time $t$ as shown in the figure. When will the chase end?

Decoding the Chase

Imagine two cars on a straight highway. One car (let's call it B) zooms past a checkpoint at a high speed but immediately starts hitting the brakes. At that exact same moment, a police car (let's call it A) sitting at the checkpoint hits the gas and starts accelerating from rest.

This is exactly what our velocity-time graph is showing us. The phrase "pass a particular point" is our cue to set our origin. We can say that at $t=0$, both particles are at $x=0$. The chase is on!

To solve this, we need the equations of motion for both particles. Let's look at the chaser, A. Its graph is a straight line passing through the origin. This means it has a constant acceleration, which we'll call $a$.

Its velocity equation is simply:

Now, look at particle B. It starts with some high initial velocity $v_0$ and has a negative slope, meaning it's decelerating. Let's call its deceleration rate $k$.

Its velocity equation is:

But we can do better! The graph shows that B's velocity hits exactly zero at $t=4\text{ s}$.

Plugging this in:

So, we can rewrite B's velocity as:

The most important feature of the graph is where the two lines cross. At $t=3\text{ s}$, both particles have the exact same velocity.

Let's equate our two velocity expressions at this instant:

This is a massive breakthrough! We've linked the deceleration of B to the acceleration of A. Now we can write B's velocity entirely in terms of $a$:

Equating Displacements

When does a chase end? It ends when the chaser's position equals the chased particle's position. Since they started at the same point, their total displacements must be equal.

In kinematics, displacement is the area under the velocity-time graph. We can find this by integrating our velocity equations from $t=0$ to the catch time $t$.

For the chaser A:

For the chased particle B:

Now, we set the displacements equal to find the time of the catch:

Notice how the acceleration $a$ appears in every single term? This means the actual value of the acceleration doesn't matter! As long as the ratio of their slopes is maintained, the catch time will be the same. We can divide the entire equation by $a$:

Let's group the $t^2$ terms together by adding $\frac{3}{2}{t}^2$ to both sides:

Since we are looking for the time after the start, $t>0$, we can safely divide by $2t$:

The chase ends exactly at 6 seconds!

There is a beautiful physical reality hidden in this math. Look at B's velocity equation: $v_B(t)=12a-3at$.

At $t=4\text{ s}$, B comes to a complete stop. But the chase doesn't end until $t=6\text{ s}$. What is B doing for those last two seconds? Its velocity becomes negative!

Particle B actually stopped, put it in reverse, and started moving backward towards A. The chaser didn't just run B down from behind; B reversed right into the chaser's path! This is why we must always trust the equations—they reveal the full story.