Geometrical Application of Calculus with Speed Two vehicles are heading for a crossroad (point $C$) and intend to pass straight through.
Vehicle $A$ is $100\,\mathrm{km}$ due North travelling at $80\,\mathrm{km}/\mathrm{hr}$ towards $C$
Vehicle $B$ is $150\,\mathrm{km}$ due East travelling at $90\,\mathrm{km}/\mathrm{hr}$ towards $C$
Show that after time $t$, the distance $D$ between them is given by:
$$D^2=100(145t^2-430t+325)$$

As Michael Hoppe mentioned, Pythagorean theorem will yield the desired function in D.
If you connect the top of the North part to the right side of the east part to create a triangle, we have, (t = time in hours):
$(100km - 80\frac{km}{hr}t)^2 + (150km - 90\frac{km}{hr}t)^2 = D^2$
Now just simplify.
