You should come up with the following notations for the distances of the car and police car such that: x is the horizontal distance of police car and y is the vertical distance of the car.
Since the police car moves to west, it means that the car moves in negative x direction, hence `(dx)/(dt) = -40` .
Since the car moves to south, it means that the car moves in negative y direction, hence `(dy)/(dt) = -50` .
Notice that the direction of police car is perpendicular to direction of the car, hence, you should use Pythagora's theorem to find the distance between the cars such that:
`d^2 = x^2 + y^2 =gt d = sqrt(x^2 + y^2)`
You need to differentiate the distance with respect to t to find the rate the distance between cars is changing.
`d'(t) = (1/(2sqrt(x^2 + y^2)))*((x^2 + y^2)')`
`d'(t) = 2(x(t)*(dx)/(dt) + y(t)*(dy)/(dt))/(2sqrt(x^2 + y^2))`
You need to substitute `1/4` for x(t), `1/2` for y(t) and -40 for `(dx)/(dt)` , -50 for `(dy)/(dt).`
`d'(t) = (-40/4 - 50/2)/(sqrt(1/16 + 1/4))`
`d'(t) = (-10 - 25)/(sqrt(5/16)) =gt d'(t) = -4*35/sqrt5`
`d'(t) = -140/sqrt5 =gt d'(t) ~~ 62.611`
Hence, evaluating the rate of change of the distance between cars yields `d'(t) ~~ 62.611 m/h` .