On a straight highway, two cars A and B are running at the same speed u = 108 km/h in the same lane.

Q: Why do we convert km/h to m/s?

Standard SI units ensure consistency, especially since the time intervals given (like 7.0 s and 1.0 s) are in seconds. Mixing hours and seconds will lead to disastrous calculation errors!

Q: In Case (a), why don't we worry about them colliding before they stop?

Because Car A (in front) is decelerating faster than Car B. From the moment B starts braking, B is always moving faster than A, meaning the gap is continuously closing until they both stop. The total distance difference is the absolute minimum gap needed.

Q: In Case (b), why is the minimum separation not just the difference in stopping distances?

Because Car A (now in the rear) has better brakes! Even though A reacts late and gets dangerously close initially, once A hits the brakes, it slows down faster than B. The critical moment is when their speeds match, not when they stop.

Q: How did you know the area in Case (b) is a triangle?

By plotting the velocity-time graphs! Car A's velocity stays constant for 1 second, forming a horizontal line, while Car B's velocity drops linearly. The point where their speeds equalize forms the third vertex, creating a perfect geometric triangle representing the extra distance.

Q: What if Car B had the stronger brakes?

If B had stronger brakes and was in the rear (Case A), we would have to check the intersection of their velocities, just like we did in Case (b) here. The math always adapts to the physical reality!

Question

On a straight highway, two cars A and B are running at the same speed $u = 108 	ext{ km/h}$ in the same lane. In the best efforts of braking, at this speed the car A can stop in $t_{sA} = 7.0 	ext{ s}$ and the car B in $t_{sB} = 10 	ext{ s}$. In an emergency when driver of the front car applies brakes, in response the driver of the rear car also has to apply brakes to avoid accident. However braking of the rear car begins after a delay $t_d = 1.0 	ext{ s}$ from the instant its driver notices the brake light signal of the front car.
(a) If car A is running ahead of car B, what should be the minimum separation between them before driver of the car A applies brake?
(b) If car B is running ahead of car A, what should be the minimum separation between them before driver of the car B applies brake?

Sigma Percentile · Accepted Answer

### The Anatomy of a Highway Emergency

Imagine you are cruising down a straight highway at a brisk $108 	ext{ km/h}$. Suddenly, the brake lights of the car ahead flare up. You slam on your brakes, but there is an inevitable human delay—a reaction time.

This problem is a classic exploration of relative motion and kinematics. It forces us to analyze not just how cars stop, but how they stop *relative to each other*.

### Setting the Stage: Deceleration

First, we must standardize our units. A speed of $108 	ext{ km/h}$ converts cleanly to $30 	ext{ m/s}$.

Next, we evaluate the braking power of both vehicles. Car A can halt in $7.0 	ext{ s}$, giving it a deceleration of $a_A = \frac{30}{7} 	ext{ m/s}^2$. Car B takes $10 	ext{ s}$ to stop, meaning its deceleration is $a_B = 3 	ext{ m/s}^2$.

**The critical insight:** Car A has significantly stronger brakes than Car B. This asymmetry completely changes the dynamics depending on who is in front!

### Case (a): The Intuitive Scenario

If Car A is leading, we have a highly dangerous situation. The car in front can stop much faster than the car behind it.

When A brakes, B continues at $30 	ext{ m/s}$ for a full $1.0 	ext{ s}$ due to reaction time. Even after B starts braking, it cannot decelerate as quickly as A. Therefore, the gap between them will continuously shrink until both vehicles come to a complete halt.

To avoid a collision, the initial separation must be greater than the difference in their total stopping distances. Car A's stopping distance is $s_A = \frac{1}{2}(30)(7) = 105 	ext{ m}$. Car B travels $30 	ext{ m}$ during the reaction time, plus $\frac{1}{2}(30)(10) = 150 	ext{ m}$ while braking, totaling $180 	ext{ m}$.

The minimum safe distance is simply $180 	ext{ m} - 105 	ext{ m} = 75 	ext{ m}$.

### Case (b): The Counter-Intuitive Trap

Now, let's flip the order. Car B is in front. B slams the brakes, and A reacts $1.0 	ext{ s}$ later.

Initially, this looks terrifying. A is hurtling forward at $30 	ext{ m/s}$ while B is already slowing down. The gap is closing rapidly! However, once A finally hits the brakes, its superior stopping power kicks in.

**The turning point:** The cars will be closest to each other at the exact instant their speeds become equal. If they don't crash before this moment, A will begin to pull away, and they are safe.

### The Geometric Masterstroke

We could use heavy algebra to find the minimum separation, but there is a far more elegant way: the velocity-time graph.

If we plot $v$ versus $t$ for both cars, the extra distance Car A travels relative to Car B is simply the area between their curves up to the moment their speeds match.

Setting their velocity equations equal, $30 - \frac{30}{7}(t - 1) = 30 - 3t$, we find they reach the same speed at $t = \frac{10}{3} 	ext{ s}$. At this instant, both are moving at $20 	ext{ m/s}$.

Looking at the graph, the area between the curves forms a perfect triangle! The base of this triangle is the $1.0 	ext{ s}$ reaction time (where A's speed is constant). The height is the drop in speed from $30 	ext{ m/s}$ to $20 	ext{ m/s}$, which is $10 	ext{ m/s}$.

The area is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2}(1)(10) = 5.0 	ext{ m}$.

This means Car A covers exactly $5.0 	ext{ m}$ more than Car B during the critical approach phase. Thus, a mere $5.0 	ext{ m}$ of initial separation is enough to prevent a crash!

This is the power of graphical analysis in physics—transforming pages of algebra into a single, beautiful geometric shape.

Visualized Solution

$u = 30 m/s, a_{A} = \frac{30}{7} m/s^{2}, a_{B} = 3 m/s^{2}$

\text{Case (a): Car A is ahead}

d_{\text{min}} = s_B - s_A = 75 \text{ m}

\text{Case (b): Car B is ahead}

v_A(t) = v_B(t) \implies t = \frac{10}{3} \text{ s}

d_{\text{min}} = \text{Area of Triangle} = 5.0 \text{ m}

The Sigma Insight: Motion in a Straight Line

The Anatomy of a Highway Emergency

Setting the Stage

Case (a)

Case (b)

The Geometric Masterstroke

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