Math Problem Statement

A police car is located 60 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 140 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 100 feet per second. How fast is the red car actually traveling along the road?

Solution

To solve this problem, let's use the Pythagorean Theorem and related rates.

Step 1: Set up the diagram and variables

  1. Let xx represent the distance along the road from the police car's position to the red car.
  2. Let y=60y = 60 feet be the perpendicular distance from the police car to the road.
  3. Let zz represent the distance from the police car to the red car, which is decreasing at a rate of dzdt=100\frac{dz}{dt} = -100 feet per second.

Given that the red car is 140 feet away from the point on the road closest to the police car, we have x=140x = 140 feet at this moment.

Step 2: Apply the Pythagorean Theorem

Using the Pythagorean Theorem: z2=x2+y2z^2 = x^2 + y^2 Differentiate both sides with respect to time tt to relate dzdt\frac{dz}{dt} and dxdt\frac{dx}{dt}: 2zdzdt=2xdxdt2z \frac{dz}{dt} = 2x \frac{dx}{dt} Simplify to solve for dxdt\frac{dx}{dt}: dxdt=zxdzdt\frac{dx}{dt} = \frac{z}{x} \cdot \frac{dz}{dt}

Step 3: Find zz at x=140x = 140

Calculate zz when x=140x = 140 feet: z=x2+y2=1402+602z = \sqrt{x^2 + y^2} = \sqrt{140^2 + 60^2} z=19600+3600=23200=152 feetz = \sqrt{19600 + 3600} = \sqrt{23200} = 152 \text{ feet}

Step 4: Substitute values into the rate equation

Now, plug in z=152z = 152, x=140x = 140, and dzdt=100\frac{dz}{dt} = -100 feet per second: dxdt=152140(100)\frac{dx}{dt} = \frac{152}{140} \cdot (-100) dxdt=152×100140=108.57 feet per second\frac{dx}{dt} = \frac{152 \times -100}{140} = -108.57 \text{ feet per second}

The red car is traveling at approximately 108.57 feet per second along the road.

Would you like more details on any step?


Here are five related questions to expand your understanding:

  1. How would the answer change if the distance between the police car and the red car were increasing instead of decreasing?
  2. What if the distance from the police car to the road were different? How would that impact the calculations?
  3. How can the related rates method be applied to other real-world motion problems?
  4. What role does the Pythagorean Theorem play in this problem?
  5. How can we interpret the negative sign in the calculated rate?

Tip: When working with related rates, always start by identifying the relationships between the variables using geometry or algebra before differentiating with respect to time.

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Math Problem Analysis

Mathematical Concepts

Pythagorean Theorem
Related Rates
Differentiation

Formulas

z^2 = x^2 + y^2 (Pythagorean Theorem)
2z * dz/dt = 2x * dx/dt (Differentiation of Pythagorean Theorem)

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 10-12