When the cruiser is 0.5 km 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 20km/h

Question

A police cruiser, approaching a right- angled intersection from the north chasing a speeding car that hass turned the corner and now is moving straight east. When the cruiser is 0.5 km 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 20km/h. if the cruiser is moving at 60km/h at the instant of measurement, what is the speed of the car?

Answer 1

Consider a right triangle with the north corner, right angle corner, and east corner.

Distance between cruiser and car can be calculated with Pythagorean theorem.

(0.5)²+(0.8)²=z²

z≈ 0.943

x²+y²=z²

dy/dt=(-60)

y=0.5

x=0.8

dz/dt=20

2x(dx/dt)+2y(dy/dt)=2z(dz/dt)

(0.8)(dx/dt)+(0.5)(-60)=0.943(20)

After calculating

dx/dt≈61.075 km/h