Hello, samsum!
An interesting problem . . .
A police cruiser, approaching a rightangled intersection $\displaystyle A$ from the north,
is chasing a speeding car that has turned the corner and is now moving straight east.
When the cruiser is 0.6 miles north of $\displaystyle A$ and the car is 0.8 miles east of $\displaystyle A$,
the police determine that the direct distance between them and the car is increasing at 20 mph.
If the cruiser is moving at 60 mph at that instant, what is the speed of the car? Code:

P *
 *
 * z
y  *
 *
 *
*         *
A x C
The police car is at $\displaystyle P,\:y$ miles north of $\displaystyle A.$
The car is at $\displaystyle C,\:x$ miles east of $\displaystyle A.$
The distance between them is: .$\displaystyle z^2 \:=\:x^2 + y^2$
Differentiate with respect to time: .$\displaystyle 2z\!\cdot\!\frac{dz}{dt} \;= \; 2x\!\cdot\!\frac{dx}{dt} + 2y\!\cdot\frac{dy}{dt}$
. . Divide by 2: .$\displaystyle z\!\cdot\frac{dz}{dt}\;=\;x\!\cdot\frac{dx}{dt} + y\!\cdot\!\frac{dy}{dt}$ [1]
We are told that: .$\displaystyle x = 0.8,\;y = 0.6,\;\frac{dz}{dt} = 20,\;\frac{dy}{dt} = 60$
. . (The distance $\displaystyle y$ is decreasing.)
When $\displaystyle x = 0.8$ and $\displaystyle y = 0.6$, the hypotenuse $\displaystyle z = 1.$
Substitute into equation [1]: .$\displaystyle 1(20) \;= \;(0.8)\!\cdot\!\frac{dx}{dt} + (0.6)(60) $
And we have: .$\displaystyle (0.8)\!\cdot\frac{dx}{dt} \:=\:56\quad\Rightarrow\quad \frac{dx}{dt} = 70$
Therefore, the speed of the car is $\displaystyle 70\text{ mph.}$
. . . BUSted!