Question

A police jeep, approaching a right-angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the jeep is $0.6km$ north of the intersection and the car is $0.8km$ to the east, the police determine with radar that the distance between them and the car is increasing at $20km{h}^{-1}$. If the jeep is moving at $60km{h}^{-1}$ at the instant of measurement, what is the speed of the car?

Answer 1

We draw a diagram of the car and jeep in the coordinate plane, using the positive x-axis as the eastbound highway and the positive y-axis as the northbound highway. Let $x$ be the position of car at time $t$.
$y=$ position of jeep at time $t$.
$s=$distance between can and jeep at time $t$.
We assume $x,y$, and $s$ to be differentiable functions of $t$.
$x=0.8km,y=0.6km,\frac{dy}{dt}=-60km{h}^{-1}$
$\frac{ds}{dt}=20km{h}^{-1}(dy/dt$ is negative because $y$ is decreasing.)
The variables are related as:
$s^2=x^2+y^2.....\left(i\right)$ (Pythagorean theorem)
Differentiate Eq. (i) with respect to $t$, we get
$\frac{d{s}^2}{dt}=\frac{d{x}^2}{dt}+\frac{d{y}^2}{dt}=\frac{d{s}^2}{ds}\cdot \frac{ds}{dt}=2s\frac{ds}{dt}$
Similarly, $\frac{d{x}^2}{dt}=\frac{d{x}^2}{dx}\cdot \frac{dx}{dt}=2x\frac{dx}{dt}$
and similarly $\frac{d{y}^2}{dt}=\frac{d{y}^2}{dy}\cdot \frac{dy}{dt}=2y\frac{dy}{dt}$
$2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$
Chain rule, $\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})=\frac{1}{\sqrt{x^2+y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})$
Evaluate, with $x=0.8km,y=0.6km,dy/dt=-60km{h}^{-1},ds/dt=20km{h}^{-1}$, and solve for $dx/dt$.
$20=\frac{1}{\sqrt{\underset{1}{\underset{}{(0.8{)}^2+(0.6{)}^2}}}}(0.8\frac{dx}{dt}+(0.6\left)\right(-60\left)\right)$
$\Rightarrow 20=0.8\frac{dx}{dt}-36\Rightarrow \frac{dx}{dt}=\frac{20+36}{0.8}=70$.
At the moment given in question, the car's speed is $70km{h}^{-1}$.