Round your answer to three decimal places. a car is traveling at 79 km/h due south at a point 4 5 kilometer north of an intersection. a police car is traveling at 63 km/h due west at a point 2 5 kilometer due east of the same intersection. at that instant, the radar in the police car measures the rate at which the distance between the two cars is changing. what does the radar gun register?

the radar gun register a rate of (-99.65 km/h)

Step-by-step explanation:#

if we assign the positive y axis to the north to south direction, positive x axis to the east to west direction and the origin at the intersection

Then the first car position yp will be

yc= y₀ - vc*t

with coordinates (0 , yc)

and the police car xp

xp = x₀ - vp*t

with coordinates (xp,0)

therefore the distance will be

r= √(xp² + yc²)

the rate of change of the distance is r'(t)=dr/dt , then

r= √(xp² + yc²)

dr/dt = 1/(2r)*(2*xp*dxp/dt+2*yp*dyc/dt) = (-1/r)*(xc*vc+yp*vp)

r'(t)=dr/dt= [-1/r(t) ]*(xc*vc+yp*vp)

at the time the radar is measuring, the distance is

r(t=0)=r₀= √(x₀² +y₀²) = √[(45 km)² +  (25 km)²]= 51.478 km

thus

r'(t=0) =  [-1/r₀ )]*(x₀*vc+y₀*vp)= -1/(51.478 km)* (45 km*79 km/h+25 km*63 km/h) = -99.65 km/h

r'(t=0) = -99.65 km/h

Step-by-step explanation:

The answer is 99.64 without negative

answer: i think it's a

step-by-step explanation:

3 qrs + 3/2 xy hope this : /

step-by-step explanation: