Question

Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 30 mph. at the same time, another car is 1/2 mile east of the intersection, driving east (away from the intersection) at an unknown speed. the officer's radar gun indicates 15 mph when pointed at the other car (that is, the straight-line distance between the officer and the other car is increasing at a rate of 15 mph). what is the speed of the other car?

Answer 1

Answer:

$v = 51.2 mph$

Explanation:

Velocity of police car

$v_p = 30 mph$ towards North

velocity of the car with respect to police car

$v_{cp} = 15 mph$

Speed of the other car is given with respect to police car so let say the speed of other car is "v"

Now the component of this speed along the line joining two cars is

$v_1 = v cos45$

similarly speed of police car along the line joining two cars is given as

$v_2 = v sin45$

now we can find the relative speed

$v_{12} = v_1 - v_2$

$15 mph = vcos45 - 30cos45$

$v = 51.2 mph$

Answer 2

In triangle ABC , using Pythagorean theorem

BC = sqrt(AB² + AC²)

r = sqrt(y² + x²)                                eq-1

taking derivative both side relative to "t"

dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))

dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))

dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)

15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)

v₂ = 51.2 m/s