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
