Solved on Jan 21, 2024

The red car is $150$ feet up a straight road from a police car $50$ feet to the side. The police radar reads the distance between the cars is decreasing at $90$ feet/second. Find the red car's actual speed along the road in feet/second.

STEP 1

Assumptions
1. The police car is stationary and located 50 feet from the road.
2. The red car is driving along the road and is initially 150 feet from the police car's perpendicular position to the road.
3. The radar reads that the distance between the police car and the red car is decreasing at 90 feet per second.
4. We are looking for the speed of the red car along the road, not the speed at which the distance between the cars is decreasing.

STEP 2

Let's denote the distance of the red car from the point on the road closest to the police car as $x$ feet. The distance between the police car and the red car is given by the radar as $z$ feet. We are given that $z$ is decreasing at a rate of 90 feet per second, so $\frac{dz}{dt} = -90$ feet per second.

STEP 3

We can use the Pythagorean theorem to relate the distances $x$ and $z$ . The distance from the police car to the road is constant at 50 feet, which we will denote as $y$ . According to the Pythagorean theorem, we have:
$z^2 = x^2 + y^2$

STEP 4

Differentiate both sides of the equation with respect to time $t$ to find the relationship between the rates of change of $x$ and $z$ .
$\frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + y^2)$

STEP 5

Apply the chain rule to differentiate both sides of the equation.
$2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$

STEP 6

Since the police car is stationary, $y$ is constant and therefore $\frac{dy}{dt} = 0$ . Simplify the equation.
$2z \frac{dz}{dt} = 2x \frac{dx}{dt}$

STEP 7

Divide both sides of the equation by $2z$ to isolate $\frac{dx}{dt}$ , which represents the speed of the red car along the road.
$\frac{dz}{dt} = \frac{x}{z} \frac{dx}{dt}$

STEP 8

We are given that $\frac{dz}{dt} = -90$ feet per second. We also know that initially, $x = 150$ feet and $y = 50$ feet. We can use these values to find the initial value of $z$ using the Pythagorean theorem.
$z^2 = x^2 + y^2$

STEP 9

Substitute the initial values of $x$ and $y$ into the equation.
$z^2 = 150^2 + 50^2$

STEP 10

Calculate the initial value of $z$ .
$z^2 = 22500 + 2500$
$z^2 = 25000$
$z = \sqrt{25000}$
$z = 158.11388300841898 \approx 158$

STEP 11

Now we can substitute the known values into the equation from STEP_7 to find $\frac{dx}{dt}$ .
$-90 = \frac{150}{158} \frac{dx}{dt}$

STEP 12

Solve for $\frac{dx}{dt}$ .
$\frac{dx}{dt} = \frac{-90 \times 158}{150}$

STEP 13

Calculate the value of $\frac{dx}{dt}$ .
$\frac{dx}{dt} = \frac{-14220}{150}$
$\frac{dx}{dt} = -94.8$

STEP 14

Since we are looking for the speed of the red car along the road, we take the absolute value of $\frac{dx}{dt}$ because speed is a scalar quantity and cannot be negative.
$Speed = |\frac{dx}{dt}|$
$Speed = |-94.8|$
$Speed = 94.8$
The actual speed (along the road) of the red car is 94.8 feet per second.

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The red car is $150$ feet up a straight road from a police car $50$ feet to the side. The police radar reads the distance between the cars is decreasing at $90$ feet/second. Find the red car's actual speed along the road in feet/second.

STEP 1

STEP 2

STEP 3

We can use the Pythagorean theorem to relate the distances $x$ and $z$ . The distance from the police car to the road is constant at 50 feet, which we will denote as $y$ . According to the Pythagorean theorem, we have:
$z^2 = x^2 + y^2$

STEP 4

Differentiate both sides of the equation with respect to time $t$ to find the relationship between the rates of change of $x$ and $z$ .
$\frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + y^2)$

STEP 5

Apply the chain rule to differentiate both sides of the equation.
$2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$

STEP 6

Since the police car is stationary, $y$ is constant and therefore $\frac{dy}{dt} = 0$ . Simplify the equation.
$2z \frac{dz}{dt} = 2x \frac{dx}{dt}$

STEP 7

Divide both sides of the equation by $2z$ to isolate $\frac{dx}{dt}$ , which represents the speed of the red car along the road.
$\frac{dz}{dt} = \frac{x}{z} \frac{dx}{dt}$

STEP 8

We are given that $\frac{dz}{dt} = -90$ feet per second. We also know that initially, $x = 150$ feet and $y = 50$ feet. We can use these values to find the initial value of $z$ using the Pythagorean theorem.
$z^2 = x^2 + y^2$

STEP 9

Substitute the initial values of $x$ and $y$ into the equation.
$z^2 = 150^2 + 50^2$

STEP 10

Calculate the initial value of $z$ .
$z^2 = 22500 + 2500$
$z^2 = 25000$
$z = \sqrt{25000}$
$z = 158.11388300841898 \approx 158$

STEP 11

Now we can substitute the known values into the equation from STEP_7 to find $\frac{dx}{dt}$ .
$-90 = \frac{150}{158} \frac{dx}{dt}$

STEP 12

Solve for $\frac{dx}{dt}$ .
$\frac{dx}{dt} = \frac{-90 \times 158}{150}$

STEP 13

Calculate the value of $\frac{dx}{dt}$ .
$\frac{dx}{dt} = \frac{-14220}{150}$
$\frac{dx}{dt} = -94.8$

STEP 14

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The red car is 150150150 feet up a straight road from a police car 505050 feet to the side. The police radar reads the distance between the cars is decreasing at 909090 feet/second. Find the red car's actual speed along the road in feet/second.

STEP 1

STEP 2

STEP 3

We can use the Pythagorean theorem to relate the distances xxx and zzz. The distance from the police car to the road is constant at 50 feet, which we will denote as yyy. According to the Pythagorean theorem, we have:z2=x2+y2z^2 = x^2 + y^2z2=x2+y2

STEP 4

Differentiate both sides of the equation with respect to time ttt to find the relationship between the rates of change of xxx and zzz.ddt(z2)=ddt(x2+y2)\frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + y^2)dtd​(z2)=dtd​(x2+y2)

STEP 5

Apply the chain rule to differentiate both sides of the equation.2zdzdt=2xdxdt+2ydydt2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}2zdtdz​=2xdtdx​+2ydtdy​

STEP 6

Since the police car is stationary, yyy is constant and therefore dydt=0\frac{dy}{dt} = 0dtdy​=0. Simplify the equation.2zdzdt=2xdxdt2z \frac{dz}{dt} = 2x \frac{dx}{dt}2zdtdz​=2xdtdx​

STEP 7

Divide both sides of the equation by 2z2z2z to isolate dxdt\frac{dx}{dt}dtdx​, which represents the speed of the red car along the road.dzdt=xzdxdt\frac{dz}{dt} = \frac{x}{z} \frac{dx}{dt}dtdz​=zx​dtdx​

STEP 8

We are given that dzdt=−90\frac{dz}{dt} = -90dtdz​=−90 feet per second. We also know that initially, x=150x = 150x=150 feet and y=50y = 50y=50 feet. We can use these values to find the initial value of zzz using the Pythagorean theorem.z2=x2+y2z^2 = x^2 + y^2z2=x2+y2

STEP 9

Substitute the initial values of xxx and yyy into the equation.z2=1502+502z^2 = 150^2 + 50^2z2=1502+502

STEP 10

Calculate the initial value of zzz.z2=22500+2500z^2 = 22500 + 2500z2=22500+2500z2=25000z^2 = 25000z2=25000z=25000z = \sqrt{25000}z=25000​z=158.11388300841898≈158z = 158.11388300841898 \approx 158z=158.11388300841898≈158

STEP 11

Now we can substitute the known values into the equation from STEP_7 to find dxdt\frac{dx}{dt}dtdx​.−90=150158dxdt-90 = \frac{150}{158} \frac{dx}{dt}−90=158150​dtdx​

STEP 12

Solve for dxdt\frac{dx}{dt}dtdx​.dxdt=−90×158150\frac{dx}{dt} = \frac{-90 \times 158}{150}dtdx​=150−90×158​

STEP 13

Calculate the value of dxdt\frac{dx}{dt}dtdx​.dxdt=−14220150\frac{dx}{dt} = \frac{-14220}{150}dtdx​=150−14220​dxdt=−94.8\frac{dx}{dt} = -94.8dtdx​=−94.8

STEP 14

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The red car is $150$ feet up a straight road from a police car $50$ feet to the side. The police radar reads the distance between the cars is decreasing at $90$ feet/second. Find the red car's actual speed along the road in feet/second.

We can use the Pythagorean theorem to relate the distances $x$ and $z$ . The distance from the police car to the road is constant at 50 feet, which we will denote as $y$ . According to the Pythagorean theorem, we have:
$z^2 = x^2 + y^2$

Differentiate both sides of the equation with respect to time $t$ to find the relationship between the rates of change of $x$ and $z$ .
$\frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + y^2)$

Apply the chain rule to differentiate both sides of the equation.
$2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$

Since the police car is stationary, $y$ is constant and therefore $\frac{dy}{dt} = 0$ . Simplify the equation.
$2z \frac{dz}{dt} = 2x \frac{dx}{dt}$

Divide both sides of the equation by $2z$ to isolate $\frac{dx}{dt}$ , which represents the speed of the red car along the road.
$\frac{dz}{dt} = \frac{x}{z} \frac{dx}{dt}$

We are given that $\frac{dz}{dt} = -90$ feet per second. We also know that initially, $x = 150$ feet and $y = 50$ feet. We can use these values to find the initial value of $z$ using the Pythagorean theorem.
$z^2 = x^2 + y^2$

Substitute the initial values of $x$ and $y$ into the equation.
$z^2 = 150^2 + 50^2$

Calculate the initial value of $z$ .
$z^2 = 22500 + 2500$
$z^2 = 25000$
$z = \sqrt{25000}$
$z = 158.11388300841898 \approx 158$

Now we can substitute the known values into the equation from STEP_7 to find $\frac{dx}{dt}$ .
$-90 = \frac{150}{158} \frac{dx}{dt}$

Solve for $\frac{dx}{dt}$ .
$\frac{dx}{dt} = \frac{-90 \times 158}{150}$

Calculate the value of $\frac{dx}{dt}$ .
$\frac{dx}{dt} = \frac{-14220}{150}$
$\frac{dx}{dt} = -94.8$