Crossing campus this afternoon, a student whose exam is later this week asked me “when you ask a real-world question on the exam and you want us to solve an ODE, can we just do it using formula we memorised from A-level physics?” Like what? “Like with one of the distance questions we might just use $v^2=u^2+2as$.” I said that if they were relying on a result we didn’t use in the module and that they hadn’t proven, this would be a problem.

In the latest Taking Maths Further podcast (Episode 19: Computer games and mechanics), we had a puzzle that we say could be answered roughly, but the precise answer 23.53 (2 d.p.) required a little calculus. On Twitter, @NickJTaylor said
https://twitter.com/NickJTaylor/status/597725017884008450

The question was: “Susan the Hedgehog runs at 20cm/s across the screen while the run button is held down. Once the run button is released, she slows down with constant deceleration of 8.5cm/s². Will she stop within 32cm more of screen?”

Taking the position to be $x$, we have constant acceleration $x^{\mathrm{\prime }\mathrm{\prime }}=-8.5$ and initial speed $x^{\prime }(0)=20$. Therefore we get, w.r.t. time $t$,

$$x^{\prime }=\int x^{\mathrm{\prime }\mathrm{\prime }}\mathrm{d}t=-8.5t+20\text{.}$$

Setting $x^{\prime }=0$ gives $t=\frac{20}{8.5}=\frac{40}{17}$ when Susan has stopped.

Now we can integrate again to get position and, since we can decide $x(0)=0$, we can omit the constant:

$$x=\int x^{\prime }\mathrm{d}t=-4.25t^2+20t\text{.}$$

Putting in $t=\frac{40}{17}$ gives

$$x=-4.25{\left(\frac{40}{17}\right)}^2+20\left(\frac{40}{17}\right)=\frac{400}{17}\approx 23.53\text{.}$$

@NickJTaylor is suggesting that we use the fact that “$v^2=u^2+2as$” or, using the notation above, $(x^{\prime }{)}^2=u^2+2ax$, where $x^{\prime }(0)=u$ and $x^{\mathrm{\prime }\mathrm{\prime }}=a$ is a constant. This is okay, and it works, but to me it still uses calculus.

To get to this, we start with $x^{\mathrm{\prime }\mathrm{\prime }}=a$, $x^{\prime }(0)=u$ and $x(0)=0$, and obtain

$$\begin{array}{}\text{(1)}& x^{\prime }& =\int x^{\mathrm{\prime }\mathrm{\prime }}\mathrm{d}t=at+u\text{;}\text{(2)}& x& =\int x^{\prime }\mathrm{d}t=\frac{1}{2}a{t}^2+ut\text{.}\end{array}$$

From (1), we rearrange for $t$ to give, for non-zero acceleration,

$$t=\frac{x^{\prime }-u}{a}\text{.}$$

Substituting this into (2), we get

$$\begin{array}{rl}x& =\frac{1}{2}a{\left(\frac{x^{\prime }-u}{a}\right)}^2+u\left(\frac{x^{\prime }-u}{a}\right)\\ & =\frac{1}{2a}(x^{\prime }-u{)}^2+\frac{1}{a}u(x^{\prime }-u)\\ & =\frac{1}{2a}((x^{\prime }{)}^2-2x^{\prime }u+u^2)+\frac{1}{a}(x^{\prime }u-u^2)\\ & =\frac{1}{2a}((x^{\prime }{)}^2–u^2)\text{.}\end{array}$$

So

$$(x^{\prime }{)}^2=u^2+2ax\text{.}$$

Setting $a=-8.5$, $u=20$ and $x^{\prime }=0$ gives

$$0=400-17x\text{,}$$

so we see $x=\frac{400}{17}\approx 23.53$.

If you are happy to accept $v^2=u^2+2as$ as a given, or to work out the area under a graph of the velocity to get displacement, then you could say there’s no calculus needed. I’d say that deriving the formula, or knowing that the area gives the displacement, uses calculus. And if you’re doing a calculus question on my exam, you should expect to have to show me the calculus.