Question

A police cruiser, 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 cruiser is $0.6$ km north of the intersection and the car is $0.8$ km to the east, the police determine with radar that the distance of the car is increasing at $20$ km/h. Suppose that the cruiser is moving at $60$ km/h at the instant of measurement.

The speed of the car is (in km/h)

• A. $70$
• B. $80$
• C. $75$
• D. $60$

Answer 1

Answer: (A)

$70$

North is taken as positive y direction and East as positive x direction

At the instant in question x=0.8, y=0.6,
$\frac{dy}{dt}=-60km/h$, $\frac{ds}{dt}=20km/h$,

$s^2=x^2+y^2$.

$2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$

Substituting the given values, we get
$20=0.8\frac{dx}{dt}-36\Rightarrow \frac{dx}{dt}=70$