# Seventy Versus One Hundred Revisited Subscribe! My latest posts can be found here:
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## Revisiting Seventy vs One Hundred

In the post Seventy Versus One Hundred, we looked at the situation of two identical cars, one doing 70, the other doing 100, slamming on their brakes at exactly the same location to avoid an obstruction in the road. The slower car stops in time, the faster doesn't, and the question was to compute the speed at impact.

The model we used there was that the two cars shed the same amount of energy in the distance they have, so we're assuming that energy is shed at a rate proportional to the distance travelled. But maybe the rate energy is shed is proportional to time, and not distance.

What happens then?

In the previous case the calculation was quite straight forward, so the answer came quickly, but this time it's less so, and we need to use the full power of the so-called "suvat" equations. Messy, yes, but it does serve to demonstrate the power of the general algebraic approach.

To make the analysis easier, let's look at starting from stationary and adding energy at a constant rate. So the energy of the car will be $E(t)=ct$ for some constant $c$. That manifests itself as kinetic energy, so we have $\frac{1}{2}mv^2=ct$, so we get a formula for velocity as a function of time:

• $v(t)=\sqrt{\frac{2c}{m}}\sqrt{t}$ (Equation 1)

We can integrate that with respect to $t$ to get the formula for distance with respect to time, made easier by thinking of this as starting from stationary, so we get:

• $d(t)=\frac{2}{3}\sqrt{\frac{2c}{m}}\;{t}^{3/2}$ (Equation 2)

Since the information we have is the speed of the cars, we need to get the distance travelled as a function of the speed, so we can rearrange Equation 1 to get this:

• $\sqrt{t}=\sqrt{\frac{m}{2c}}\;{v}$

We can cube that to get ${t}^{3/2}$:

• ${t}^{3/2}=\left({\frac{m}{2c}}\right)^\frac{3}{2}\;{v^3}$

Substitute into Equation 2 and simplify:

• $d={\frac{m}{3c}}\;v^3$ (Equation 3)

So if $u$ is the initial speed of the slower car, the distance to the obstruction is ${\frac{m}{3c}}u^3$. If $v$ is the initial speed of the faster car, then the distance it will travel before coming to a stop is ${\frac{m}{3c}}v^3$, so the distance beyond the obstruction that it would travel is ${\frac{m}{3c}}(v^3-u^3)$.

So for that distance of travel, how fast would it be going? That's easily obtained from Equation 3 and we get:

• Speed at collision = $(v^3-u^3)^{1/3}$

Note that the constants of $c$ and $m$ all cancel, which is as it should be.

So with initial speeds of 70 and 100 (mph or kph, doesn't really matter) the speed at collision is about 87. Alternatively, with initial speeds of 30 and 40, the speed at collision is over 33.

So if you're doing 40 in a 30 zone and something crops up at a distance where you could stop if travelling at the speed limit, by the time you hit it, you'll still be travelling faster than the limit.

That's rather sobering.

## Comparing models

So which model is correct? The short answer is that we don't know. However, Chris Robbins (@Grallator on Twitter) suggests that the force applied by the brake pads on the brake disc is going to be roughly constant, and so the work done, which in this case is the energy shed, is the force times the distance over which the force is applied.

That means that the energy shed will be a function of the distance travelled, so the calculation done in Seventy Versus One Hundred would be correct.

But sometimes the efficiency of the brakes varies under temperature, so in that case perhaps we need to look at how fast the brakes can radiate the energy, and so we're back to energy shed per time, as we've calculated here.

Most likely it's a mixture of the two, but regardless, what's clear is that the stopping distance, and the speed at collision for these scenarios, is soberingly large.

## References and credits

As always, I'm indebted to the input from and interactions with a few people. The problem was first mentioned to me by Ben Sparks, who has done a video on this with Numberphile.

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