# At that instant, the radar in the police car measures the rate at which the distance between the 2 cars is changing. What does the radar given register?A car is traveling at 50mph due south at a...

At that instant, the radar in the police car measures the rate at which the distance between the 2 cars is changing. What does the radar given register?

A car is traveling at 50mph due south at a point 1/2 mile north of an intersection point. A police car is traveling at 40 mph due west at a point 1/4 mile due east of the same intersection point. At that instant, the radar in the police car measures the rate at which the distance between the 2 cars is changing. What does the radar given register?

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### 1 Answer

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` .**