Answer to Question #155233 in Calculus for rachel

Question

A police car is located 70 feet to the side of a straight road.

A red car is driving along the road in the direction of the police car and is 100 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 70 feet per second. How fast is the red car actually traveling along the road?

Answer 1

A police car is located 70 feet to the side of a straight road.

A red car is driving along the road in the direction of the police car and is 100 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 70 feet per second. How fast is the red car actually traveling along the road?

solution:

The police car and the red car at the two vertices of a right angle triangle with leg of 100feet and 70feet. The line of sight between them is the hypotenuse , S.

The information on the radar can be expressed as ${dS \over dt}=- 70feet/sec$

Let r be the speed of the Red car which we want to find.

We'll see that the length of the Red car's leg is decreasing and the relation is given by

$100-rt$

where rt is the speed of the red car multiply by time, and this gives us the distance with which the red car's leg is decreasing as time passes.

The hypotenuse of the right angle triangle formed at time t=0 can be found from;

$S^2=30^2+100^2$

$\because$ $S=104.40feet$

The distance between the two car has the relationship

$S^2=30^2+(100-rt)^2$

Taking the derivative of both sides with respect to t

$2S{dS \over dt}=0 - 2r(100-rt)$

$2S{dS \over dt}=- 2r(100-rt)$

The expression is true for all values of t.

Now, let's use the values we have for t=0

$2(104.40)(-70)=-2(r)(100-r(0))$

$-14616=-200r$

$r= 73.08feet/sec$

Note that this value is larger than speed measure on the radar. This should be expected because both cars are in motion