Car Catastrophe

Classical Mechanics Level 2

Two identical cars are traveling on a road. The first car is travelling at $\SI[per-mode=symbol]{25}{\meter\per\second}$ . The second car is overtaking the first at $\SI[per-mode=symbol]{30}{\meter\per\second}$ .

When they are next to each other, they both notice that the road is blocked ahead by a crashed truck. They apply their brakes at the same time, and the brakes apply an identical force to slow the cars down.

If the first car stops just before the crashed truck, how fast (in m/s) is the second car traveling when it hits the truck? Submit your answer to 2 decimal places.

The answer is 16.58.

1 solution

Nicholas James
Mar 14, 2017

It is possible to solve this question by using the uniform motion equations, and inventing either a figure for the distance to the truck, or inventing a figure for the acceleration.

A better way is to use energy. The formula for kinetic energy is $KE=\frac{1}{2}mv^2$ . At the start, the first car has kinetic energy $E_1$ :

$E_1=\frac{1}{2}m(25)^2$

Just before the truck, it has lost all its kinetic energy thanks to the work done by its brakes.

At the start, the second car has energy $E_2$ :

$E_1=\frac{1}{2}m(30)^2$

By the time it reaches the truck, it will have lost the same amount of energy as the first car, as they have both been acted on by the same force over the same distance. It will have $E_{final}$ kinetic energy remaining:

$E_{final}=\frac{1}{2}m(30)^2-\frac{1}{2}m(25)^2\\ E_{final}=\frac{1}{2}m(275)$

We want to find the final velocity, $v$ , so:

$\frac{1}{2}mv^2=\frac{1}{2}m(275)\\ v^2=275\\ v=\boxed{16.58...}$

This is nearly $40mph$ , so the initial difference in velocities has been vastly magnified. This is because kinetic energy is proportional to $v^2$ , not simply $v$ .

The formula $v_{f}^{2} - v_{i}^{2} = 2ad$ does the trick as well. Since $2ad$ is the same for both cars, we have that $0^{2} - 25^{2} = v^{2} - 30^{2} \Longrightarrow v^{2} = 275$ .

Brian Charlesworth - 4 years, 2 months ago

:) It's almost as if all these formulas are describing the same things!

Nicholas James - 4 years, 2 months ago

Yes, they do seem to in this instance. The work done by the brakes in slowing each car is $Fd = mad$ , and this corresponds to the loss in kinetic energy, i.e,

$\dfrac{1}{2}mv_{f}^{2} - \dfrac{1}{2}mv_{i}^{2} = mad \Longrightarrow v_{f}^{2} - v_{i}^{2} = 2ad$ .

P.S.. It was surprising how fast the second car crashed into the truck, given that the initial speeds differed by only 5 m/s. I guess that's why we should stick to the speed limits. :)

Brian Charlesworth - 4 years, 2 months ago

@Brian Charlesworth – It's nice, actually, because that's the one of the 4 suvat equations that doesn't fall out of a velocity time graph.

Nicholas James - 4 years, 2 months ago

@Brian Charlesworth – I also was surprised how fast the second car crashed. I spent about a minute checking my math because intuition suggested to me that it must have slowed down more by then. I appreciate instances in which math disproves intuition.

Tristan Goodman - 2 years, 3 months ago

Car Catastrophe

The answer is 16.58.

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