Question

A car travels down a highway at 45 m/s. An observer stands 200 m from the highway.

How fast is the distance from the observer to the car increasing when the car passes in front of the observer?

How fast is the distance increasing 40 s later?

Answer 1

check the picture below.

so, the left-side is when the car is right in front of the observer, at that point, notice the y-distance, is pancaked, so is 0, but increasing, we know dy/dt is 45 m/s, and notice the "r" distance, is the same as the "x" distance, keeping in mind that whilst "r" is increasing or moving about, "x" is static, is a constant.


`\bf \stackrel{\textit{pythagorean theorem}}{r^2=x^2+y^2}\implies 2r\cfrac{dr}{dt}=\stackrel{constant}{0}+2y\cfrac{dy}{dt} \\\\\\ 2r\cfrac{dr}{dt}=0+2(0)(45)\implies \cfrac{dr}{dt}=0`

now, 40 seconds later, since the car is going at 45 m/s, in 40 seconds it has covered 40 * 45 meters, or 1800, so y = 1800, x = 200, what is "r"?


`\bf r^2=x^2+y^2\implies r=√(200^2+1800^2)\implies r=√(3280000) \\\\\\ r=√(200^2\cdot 82)\implies r=200√(82)\\\\ -------------------------------\\\\ 2r\cfrac{dr}{dt}=0+2y\cfrac{dy}{dt}\implies \cfrac{dr}{dt}=\cfrac{y(dy)/(dt)}{r}\quad \begin{cases} r=200√(82)\\ y=1800\\ (dy)/(dt)=45 \end{cases} \\\\\\ \cfrac{dr}{dt}=\cfrac{1800\cdot 45}{200√(82)}\implies \cfrac{dr}{dt}=\cfrac{405}{√(82)}\implies \cfrac{dr}{dt}=\cfrac{405√(82)}{82}`