Answer:
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