A police car is located 60ft to the side of a straight road.
A red car is driving in the direction of the police car and is 90 ft up the road from the location of the police car. The police radar read that the distance between the red car and the police car is decreasing at a rate of 95 ft per second. How fast is the red car actually traveling along the road ?
If the red car is x feet away from the point of closest approach, then the distance z from the car to the cop is
z^2 = x^2 + 60^2
at x=90, z = 30√13
Now you have
z dz/dt = x dx/dt
so plug in your numbers
We are given:
- The police car is located 60 ft to the side of a straight road.
- A red car is driving in the direction of the police car and is 90 ft up the road from the location of the police car.
- The police radar reads that the distance between the red car and the police car is decreasing at a rate of 95 ft per second.
Let's denote:
- The distance between the red car and the police car as "d" at any given time.
- The distance traveled by the red car along the road as "x" at any given time.
We know that the distance between the red car and the police car is decreasing at a rate of 95 ft per second. Therefore, the rate of change of "d" with respect to time, denoted as "dd/dt" is equal to -95 ft/s.
We are interested in finding the rate at which the red car is actually traveling along the road, which is the rate of change of "x" with respect to time, denoted as "dx/dt".
Now, let's consider the diagram:
---------- 90 ft ---------- Red Car (x ft)
--------------------------------
|
|
|
60 ft
|
------------------------------------------------------------------------------------------------
Police Car
0 ft
We can see that the situation forms a right-angled triangle between the red car, the police car, and the side of the road.
Using the Pythagorean theorem for right-angled triangles, we know that:
d^2 = x^2 + 60^2
Taking the derivative of both sides with respect to time, we get:
2d * dd/dt = 2x * dx/dt
Now we can substitute the given values:
2(90) * (-95) = 2x * (dx/dt)
We can simplify this equation to find dx/dt:
-17100 = 2x * (dx/dt)
Dividing both sides by 2x:
(dx/dt) = -17100 / (2x)
Since we want to find the rate at which the red car is traveling along the road, dx/dt, we need to find the value of x. To do this, we can use the Pythagorean theorem:
d^2 = x^2 + 60^2
Plugging in the values:
90^2 = x^2 + 60^2
8100 = x^2 + 3600
x^2 = 4500
x = sqrt(4500)
x = 67.082
Now we can substitute the value of x into the equation for dx/dt:
(dx/dt) = -17100 / (2 * 67.082)
(dx/dt) ≈ -127.99 ft/s
Therefore, the red car is actually traveling along the road at a rate of approximately 127.99 ft/s.
To determine the speed at which the red car is traveling along the road, we can apply the concept of related rates. This involves differentiating the given information and using it to find the desired rate.
Let's start by identifying the relevant variables:
- The distance between the red car and the police car is decreasing at a rate of 95 ft per second.
- The distance between the police car and the red car (along the road) is 90 ft.
Now, let's differentiate the given information with respect to time (t):
d(distance between red car and police car)/dt = -95 ft/s (negative sign indicates decreasing distance)
We are interested in finding the velocity of the red car along the road, which can be represented as:
d(distance of red car along the road)/dt = ?
Since the distance between the red car and the police car is equal to the hypotenuse of a right triangle (formed by the police car, red car, and the road), we can use the Pythagorean theorem.
Applying the Pythagorean theorem:
(distance of red car along the road)^2 + (distance between the police car and the red car)^2 = (total distance)^2
Let's assign variables to the distances:
Let x be the distance of the red car along the road (in ft).
Plugging in the known values:
x^2 + 90^2 = (total distance)^2
Differentiating the equation with respect to time (t):
2x*(dx/dt) = 0 ---> 2x*(dx/dt) = 0
Now, we need to find the value of dx/dt (which represents the rate at which the red car is traveling along the road).
Rearranging the equation to solve for dx/dt:
2x*(dx/dt) = 0
dx/dt = 0 / 2x
dx/dt = 0
Therefore, the rate at which the red car is traveling along the road is 0 ft/s.
Explanation:
The red car is not actually moving along the road. Although the distance between the red car and the police car is decreasing at a rate of 95 ft/s, we are told that the red car is positioned 90 ft up the road. However, since the police car is located to the side of the road and not on the road itself, it means that the red car is stationary in relation to the road. Therefore, the red car's speed along the road is 0 ft/s.