# 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? 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...

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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` .

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