A question was recently posed to Click & Clack Talk Cars (http://www.greatfallstribune.com/story/life/2014/08/07/click-clack-rainy-day-raises-physics-question/13750681/). The topic is rain hitting an automobile windshield and the basic question asked is, "Are more raindrops hitting the windshield while moving." Their response is that this problem "is a very straightforward calculus problem" and that "...yes, your windshield does get hit with more raindrops per second if you are moving forward."
They state that they are incapable of solving the calculus problem and then provide an analogy which seems to support their conclusion.
My request: Would a calculus person please explain the steps involved in solving this "straightforward calculus" problem?
Since all quantities involved are constant in time and space there is no calculus needed to tackle this problem.
We may assume that there are $N\gg1$ equal-sized rain drops per unit of volume, and that all of these drops fall with the same velocity $${\bf r}=(r_1,r_2,-r_3),\quad r_3>0\ .$$ The order of magnitude is about $12\>$mph for $r_3$.
Consider a test surface $S$ with unit normal ${\bf n}$. When ${\bf n}$ is parallel to ${\bf r}$ the rain drops hitting $S$ in the next second fill a cylindrical volume with base $S$ and height $|{\bf r}|$. The number of these drops is therefore given by $$N\,{\rm area}(S)\,|{\bf r}|\ .$$ When ${\bf n}$ is tilted by an angle $\alpha$ with respect to ${\bf r}$ this number drops to $$N\,{\rm area}(S)\,|{\bf r}|\cos\alpha=N\,{\rm area}(S)\>{\bf r}\cdot{\bf n}\ .\tag{1}$$ The car moves in $x$-direction at velocity $${\bf v}=(v,0,0),\qquad v\geq0\ ,$$ and its windshield is inclined by an angle $\theta$, $\>0\leq\theta<{\pi\over2}$, with respect to the vertical. It follows that the inward unit normal of the windshield is given by $${\bf n}=(-\cos\theta,0,-\sin\theta)\ .$$ Now the relative velocity of the rain drops with respect to the moving windshield $S$ is given by $${\bf p}:={\bf r}-{\bf v}=(r_1-v,\>r_2,\>-r_3)\ .$$ In order to compute the number $\Phi$ of rain drops hitting $S$ per second we have to replace ${\bf r}$ in $(1)$ by ${\bf p}$ and so obtain $$\Phi=N\,{\rm area}(S)\>({\bf r}-{\bf v})\cdot{\bf n}=N\,{\rm area}(S)\>\bigl((v-r_1)\cos\theta+r_3\sin\theta\bigr)\ .$$ Here the right hand side is an increasing function of $v$.