Question

A police car is approaching a right-angled intersection from the north at 35 mph, and a speeding car is heading east at 30 mph. When the police car is 0.75 miles north of the intersection and the speeding car is 1 mile to the east, what is the rate that the distance between the police car and car is changing?

Answer 1

Answer:

The answer is increasing by 3 mph

Step-by-step explanation:

Answer 2

Answer:

The rate that the distance between the police car and car is changing = 3 m/s

Step-by-step explanation:

The given speed of the police car = 35 mph

The speed of the speeding car = 30 mph

The position of the police car = 0.75 miles north

The position of the speeding car = 1 miles east

The distance of the police car from the speeding car, d = √(0.75² + 1²) = 1.25

d² = x² + y²

2d·dd/dt = 2x·dx/dt + 2·y·dy/dt

Where;

dx/dt = 30 mph

dy/dt = -35 mph

x = 1

y = 0.75

Substituting gives;

2×1.25 ×dd/dt = 2×1×30 - 2×0.75×35

2×1.25 ×dd/dt = 7.5

dd/dt = 7.5/2.5 = 3

dd/dt = 3 m/s

The rate that the distance between the police car and car is changing = dd/dt = 3 m/s