Animated Solution for Physics - Kinematics: Comprehension Passage

Imagine you are standing on a hill overlooking a long, straight highway. Somewhere along this highway, there is a narrow bridge. Two cars, let's call them Car A (the leader) and Car B (the follower), are cruising down the road.

The graph provided in the problem is not just a collection of abstract lines; it is a vivid story of these two cars. By carefully decoding the turning points on this graph, we can reconstruct their entire journey, calculate their speeds, and even measure the length of the bridge without ever stepping foot on it.

Let's start by looking at the very beginning of the graph. From $t=0$ to $t=10\text{ s}$, the graph is perfectly flat. The separation between the two cars is a constant $s_1=500\text{ m}$.

What does a constant separation mean physically? It means both cars are traveling at the exact same speed. Since they are both on the open road, they are both moving at the road speed, $v_1$. As long as their speeds are identical, the distance between them cannot change.

Entering the Bridge

At exactly $t_1=10\text{ s}$, the graph takes a sharp dive. The separation between the cars starts to decrease. Why would the gap between them suddenly shrink?

This happens because the front car, Car A, has just hit the narrow bridge. Due to the narrow passage, Car A is forced to slow down to a new speed, $v_2$. However, Car B is still $500\text{ m}$ behind on the open road, blissfully zooming along at the faster speed $v_1$. Because Car B is moving faster than Car A, it starts catching up, and the separation decreases.

This catching-up phase continues until $t_2=30\text{ s}$. At this exact moment, the separation stops decreasing and becomes constant again at $s_2=200\text{ m}$.

Why did it become constant again? Because Car B has finally reached the bridge and slowed down to $v_2$ as well! Now both cars are on the bridge, moving at the same slower speed, so their separation is locked at $200\text{ m}$.

This gives us a brilliant piece of information. When Car A entered the bridge at $10\text{ s}$, Car B was exactly $500\text{ m}$ behind it. Car B reached the bridge at $30\text{ s}$. This means Car B traveled that $500\text{ m}$ distance on the road in exactly $20\text{ s}$ ($30\text{ s}-10\text{ s}$).

We can now easily calculate the road speed:

Now, let's shift our focus to Car A during that same $20\text{ s}$ window (from $10\text{ s}$ to $30\text{ s}$).

During this time, Car A was driving on the bridge. How far did it get? Well, when Car B finally arrives at the start of the bridge at $30\text{ s}$, the gap between them is $200\text{ m}$. This implies that Car A managed to travel exactly $200\text{ m}$ along the bridge before Car B even entered it.

So, Car A traveled $200\text{ m}$ on the bridge in $20\text{ s}$. We can calculate the bridge speed:

Finding the Bridge Length

Finally, we need to find the total length of the bridge. Let's look at the graph one last time. At $t_3=80\text{ s}$, the separation starts to increase.

Why? Because Car A has reached the end of the bridge and accelerated back to the faster road speed $v_1$. Car B is still stuck on the bridge at the slower speed $v_2$, so Car A starts pulling away.

This tells us exactly how long Car A was on the bridge. It entered at $t_1=10\text{ s}$ and exited at $t_3=80\text{ s}$. Therefore, Car A spent a total of $70\text{ s}$ ($80\text{ s}-10\text{ s}$) driving on the bridge.

Since we know Car A was traveling at $10\text{ m/s}$ while on the bridge, the total length of the bridge $L$ is simply:

In physics, it is always deeply satisfying to verify our results using an alternative perspective. Let's look at Car B.

Car B entered the bridge at $t_2=30\text{ s}$. When did it exit? The graph shows the separation becomes constant again at $t_4=100\text{ s}$, meaning Car B has also exited the bridge and sped back up to $v_1$.

Car B spent from $30\text{ s}$ to $100\text{ s}$ on the bridge, which is exactly $70\text{ s}$.

The logic holds perfectly. By mapping the mathematical turning points of a graph to the physical events of the real world, we unlocked the entire kinematic puzzle.