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.
[tex]\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[/tex]
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"?
[tex]\bf r^2=x^2+y^2\implies r=\sqrt{200^2+1800^2}\implies r=\sqrt{3280000}
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r=\sqrt{200^2\cdot 82}\implies r=200\sqrt{82}\\\\
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2r\cfrac{dr}{dt}=0+2y\cfrac{dy}{dt}\implies \cfrac{dr}{dt}=\cfrac{y\frac{dy}{dt}}{r}\quad
\begin{cases}
r=200\sqrt{82}\\
y=1800\\
\frac{dy}{dt}=45
\end{cases}
\\\\\\
\cfrac{dr}{dt}=\cfrac{1800\cdot 45}{200\sqrt{82}}\implies \cfrac{dr}{dt}=\cfrac{405}{\sqrt{82}}\implies \cfrac{dr}{dt}=\cfrac{405\sqrt{82}}{82}[/tex]
