Animated Solution for Physics - Kinematics: An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e, 120 km/h, the stopping distance will be

Imagine you are cruising down the highway at a comfortable $60\text{ km/h}$. Suddenly, an obstacle appears, and you slam on the brakes. Your car skids and finally comes to a complete halt after covering a distance of $20\text{ m}$.

Now, let's raise the stakes. What if you were driving twice as fast, at $120\text{ km/h}$? Intuition might suggest that doubling your speed would simply double your stopping distance to $40\text{ m}$. But physics tells a very different, and much more dangerous, story. Let's dive into the mathematics of motion to uncover the truth.

To understand how stopping distance relates to initial speed, we need a mathematical tool that connects velocity, acceleration, and displacement. Since we don't know the time it takes to stop, the third equation of motion is our perfect candidate:

Here, $u$ is the initial speed, $v$ is the final speed, $a$ is the acceleration, and $s$ is the stopping distance. Because the car is coming to a stop, its final velocity $v$ is exactly $0$. Furthermore, the brakes apply a retarding force, meaning the acceleration $a$ is acting opposite to the motion. We can represent this retardation as a negative value, let's call it $-a$.

Substituting these conditions into our equation, we get:

By rearranging the equation, we can isolate the stopping distance $s$:

This simple equation is incredibly revealing. Assuming the braking force (and therefore the retardation $a$) remains constant regardless of how fast you are going, the denominator $2a$ is just a fixed number. This leaves us with a profound relationship:

This means that any change in your speed will have an exponentially larger impact on your stopping distance.

Let's apply this proportionality to our two scenarios. We can set up a ratio comparing the new stopping distance $s_2$ to the original stopping distance $s_1$:

We know our original speed $u_1$ was $60\text{ km/h}$ and our new speed $u_2$ is $120\text{ km/h}$. Plugging these numbers in:

Finally, multiplying both sides by $20$, we find our new stopping distance:

So, doubling your speed doesn't just double your stopping distance—it quadruples it! A car travelling at $120\text{ km/h}$ will need a massive $80\text{ m}$ to stop, compared to just $20\text{ m}$ at $60\text{ km/h}$. This non-linear relationship is exactly why high-speed collisions are so devastating. The correct answer is (d) 80 m.