Explanation:
it was not clear if an average change rate would be sufficient, or if you needed an immediate change rate (as I also don't know if you covered derivatives already or not).
so, it would be helpful, if you could put a message to an answer that was not giving you what you need.
so, here now an answer for an immediate change rate (hopefully that is what you need) :
we have a right-angled triangle.
the direct line of sight (the direct distance between police and red car) is the Hypotenuse (the baseline opposite of the 90° angle).
the 50 ft and 180 ft are the legs.
Pythagoras gives us the length of the Hypotenuse :
Hypotenuse² = 50² + 180² = 2500 + 32400 = 34,900
Hypotenuse = sqrt(34900) = 186.8154169... ft
in general terms let's say x is the distance of the cop to the road, y is the distance on the road to the crossing point with the distance cop to road, and z is the line of sight distance between the red car and the cop (the Hypotenuse).
x² + y² = z²
now, the first derivative of distance is the change of distance = speed.
then dy/dt (= y') is how fast the car is traveling down the road. dx/dt (= x') is how fast the cop is traveling toward the road. and dz/dt (= z') is how fast the distance between the cop and the car is changing.
now, we take the derivative of our equation
x² + y² = z² with respect to time, variable by variable :
d(x² + y² = z²)/dt =
dx²/dx × dx/dt + dy²/dy × dy/dt = dz²/dz × dz/dt
that gives us the equation
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
x(dx/dt) + y(dy/dt) = z(dz/dt)
from the problem we know x (50 ft), y (180 ft), dz/dt (85 ft/s). we calculated z (the Hypotenuse = sqrt(34900), and since the cop is not moving, we know dx/dt = 0.
and we get
50ft×0ft/s + 180ft×(y') = sqrt(34900)ft×(85)ft/s
we solve for y' (the speed of the car on the road)
y' = sqrt(34900)×85/180 = 88.21839132... ft/s
≈ 88.22 ft/s
and now here the difference for an average change rate over the unrevealed of 1 second :
the radar measured the change of the distance (Hypotenuse) from 1 second ago to now.
so, 1 second ago, the distance was
186.8154169... + 85 = 271.8154169... ft
the 50 ft leg stays the same, but the 180 ft leg was (again via Pythagoras)
271.8154169...² = 50² + leg²
leg² = 271.8154169...² - 50² = 71,383.62088...
leg = 267.1771339... ft
so, the red car traveled
267.1771339... - 180 = 87.1771339... ft/s
as you can see, it is close, but there has to be a difference, as the average change rate is only an approximation to the immediate change rate.