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?

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

may i see the picture? it would a lot, xox

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?