Question

a police car is located 30 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 190 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 90 feet per second. how fast is the red car actually traveling along the road? the actual speed (along the road) of the red car is feet per second question help: video message instructor submit question jump to answer

Explanation:

Step1: Define Variables

Let \( x \) be the distance of the red car from the point on the road closest to the police car, \( y \) be the distance between the police car and the red car, and \( 30 \) be the perpendicular distance from the police car to the road (constant). By Pythagoras, \( y^2 = x^2 + 30^2 \).

Step2: Differentiate with Respect to Time

Differentiate both sides: \( 2y \frac{dy}{dt} = 2x \frac{dx}{dt} \), simplify to \( y \frac{dy}{dt} = x \frac{dx}{dt} \).

Step3: Find \( y \)

Given \( x = 190 \), \( y = \sqrt{190^2 + 30^2} = \sqrt{36100 + 900} = \sqrt{37000} = 10\sqrt{370} \approx 192.35 \).

Step4: Substitute Known Values

We know \( \frac{dy}{dt} = -90 \) (decreasing), \( x = 190 \), \( y \approx 192.35 \). Substitute into \( y \frac{dy}{dt} = x \frac{dx}{dt} \):

\( 192.35(-90) = 190 \frac{dx}{dt} \)

Solve for \( \frac{dx}{dt} \):

\( \frac{dx}{dt} = \frac{192.35(-90)}{190} \approx -91.17 \) (negative sign indicates direction, speed is magnitude, but wait, actually, let's do exact:

From \( y^2 = x^2 + 900 \), so \( y = \sqrt{x^2 + 900} \). Then \( \frac{dy}{dt} = \frac{x}{\sqrt{x^2 + 900}} \frac{dx}{dt} \). Wait, no, earlier differentiation: \( 2y \frac{dy}{dt} = 2x \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{y}{x} \frac{dy}{dt} \). Wait, I had a sign error. Since \( x \) is decreasing (red car moving towards the point), \( \frac{dx}{dt} \) is negative, but \( \frac{dy}{dt} \) is negative (distance decreasing). Let's re-express:

Let \( x \) be the distance from red car to the closest point (so as red car moves towards the point, \( x \) decreases, \( \frac{dx}{dt} \) is negative). The distance \( y \) between police and red car: \( y^2 = x^2 + 30^2 \). Differentiate: \( 2y \frac{dy}{dt} = 2x \frac{dx}{dt} \implies y \frac{dy}{dt} = x \frac{dx}{dt} \). We need \( \frac{dx}{dt} \) (rate of change of \( x \), but the speed along the road is \( |\frac{dx}{dt}| \) if \( x \) is decreasing, but actually, the speed of the red car is \( -\frac{dx}{dt} \) (since \( \frac{dx}{dt} \) is negative when moving towards the point). Wait, let's plug in values:

\( x = 190 \), \( \frac{dy}{dt} = -90 \) (decreasing), \( y = \sqrt{190^2 + 30^2} = \sqrt{36100 + 900} = \sqrt{37000} = 10\sqrt{370} \approx 192.354 \)

Then \( 192.354(-90) = 190 \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{192.354(-90)}{190} \approx -91.17 \)

But the speed along the road is the magnitude of the velocity, but actually, the red car's speed is \( -\frac{dx}{dt} \) because \( \frac{dx}{dt} \) is negative (x is decreasing). Wait, no: if \( x \) is the distance from the red car to the point closest to the police car, then as the red car moves towards that point, \( x \) decreases, so \( \frac{dx}{dt} \) is negative. The speed of the red car is \( |\frac{dx}{dt}| \), but actually, in the equation, we have \( \frac{dx}{dt} \) as the rate of change of \( x \), and the speed along the road is \( -\frac{dx}{dt} \) (since \( \frac{dx}{dt} \) is negative, speed is positive). Wait, let's re-express the differentiation correctly.

Let me define \( x \) as the distance from the red car to the point on the road closest to the police car, so when the red car moves towards the police car's closest point, \( x \) decreases, so \( \frac{dx}{dt} \) is negative (velocity component along x). The distance \( y \) between police and red car: \( y^2 = x^2 + 30^2 \). Differentiate:

\( 2y \frac{dy}{dt} = 2x \frac{dx}{dt} \implies y \frac{dy}{dt} = x \frac{dx}{dt} \)

We know \( \frac{dy}{dt} = -90 \) (since y is decreasing), \( x = 190 \), \( y = \sqrt{190^2 + 30^2} = \sqrt{36100 + 900} = \sqrt{37000} = 10\sqrt{370} \approx 192.354 \)

So:

\( 192.354(-90) = 190 \frac{dx}{dt} \)

\( \frac{dx}{dt} = \frac{192.354(-90)}{190} \approx -91.17 \)

But the speed along the road is the magnitude of the velocity, which is \( |\frac{dx}{dt}| \), but wait, no—actually, the red car's speed along the road is \( -\frac{dx}{dt} \) because \( \frac{dx}{dt} \) is negative (x is decreasing, so the speed is the rate at which x is decreasing, which is \( -\frac{dx}{dt} \)). Wait, let's check units: \( \frac{dx}{dt} \) has units of feet per second, negative means direction towards the point. So the speed is \( -\frac{dx}{dt} \approx 91.17 \). Wait, but let's do exact calculation:

\( y = \sqrt{190^2 + 30^2} = \sqrt{36100 + 900} = \sqrt{37000} = 10\sqrt{370} \)

Then from \( y \frac{dy}{dt} = x \frac{dx}{dt} \), we can solve for \( \frac{dx}{dt} \):

\( \frac{dx}{dt} = \frac{y}{x} \frac{dy}{dt} \)

Plug in:

\( \frac{dx}{dt} = \frac{10\sqrt{370}}{190} (-90) = \frac{\sqrt{370}}{19} (-90) \approx \frac{19.235}{19} (-90) \approx 1.0124 (-90) \approx -91.12 \)

So the speed along the road is the absolute value? Wait, no—actually, the red car's velocity along the road is \( \frac{dx}{dt} \), but since it's moving towards the point, \( \frac{dx}{dt} \) is negative, and the speed is the magnitude, but the question says "how fast is the red car actually traveling along the road", which is the speed, so positive value. Wait, but maybe I made a mistake in the sign. Let's think again:

When the red car moves towards the police car's closest point, \( x \) decreases, so \( \frac{dx}{dt} < 0 \). The distance \( y \) between them also decreases, so \( \frac{dy}{dt} < 0 \). From the equation \( y \frac{dy}{dt} = x \frac{dx}{dt} \), both sides are negative (since \( y > 0 \), \( \frac{dy}{dt} < 0 \); \( x > 0 \), \( \frac{dx}{dt} < 0 \)), so the equation holds. To find the speed (magnitude of velocity) along the road, we take \( |\frac{dx}{dt}| \), but actually, the velocity along the road is \( \frac{dx}{dt} \), but speed is the absolute value. Wait, no—the question says "how fast is the red car actually traveling along the road", which is the speed, so we need to find \( |\frac{dx}{dt}| \), but let's check the calculation again.

Wait, maybe I mixed up \( \frac{dx}{dt} \) and the speed. Let's re-express the problem:

Let \( v \) be the speed of the red car along the road (positive, since it's speed). Then \( \frac{dx}{dt} = -v \) (because \( x \) is decreasing at rate \( v \)).

From the differentiation: \( y \frac{dy}{dt} = x \frac{dx}{dt} \)

We have \( \frac{dy}{dt} = -90 \) (distance decreasing), \( x = 190 \), \( y = \sqrt{190^2 + 30^2} \)

So:

\( y (-90) = 190 (-v) \)

Simplify:

\( 90y = 190v \)

\( v = \frac{90y}{190} = \frac{9y}{19} \)

Calculate \( y = \sqrt{190^2 + 30^2} = \sqrt{36100 + 900} = \sqrt{37000} = 10\sqrt{370} \approx 192.354 \)

Then \( v = \frac{9 \times 192.354}{19} \approx \frac{1731.186}{19} \approx 91.115 \approx 91.12 \)

Wait, this is the correct approach: since \( \frac{dx}{dt} = -v \) (x decreasing at rate v), and \( \frac{dy}{dt} = -90 \) (y decreasing at rate 90), then substituting into \( y \frac{dy}{dt} = x \frac{dx}{dt} \):

\( y (-90) = x (-v) \implies 90y = xv \implies v = \frac{90y}{x} \)

Ah! Here's the mistake earlier: I had \( \frac{dx}{dt} = \frac{y}{x} \frac{dy}{dt} \), but if \( \frac{dx}{dt} = -v \) and \( \frac{dy}{dt} = -90 \), then:

\( y (-90) = x (-v) \implies 90y = xv \implies v = \frac{90y}{x} \)

That's the correct relation. So let's recalculate:

\( y = \sqrt{190^2 + 30^2} = \sqrt{36100 + 900} = \sqrt{37000} = 10\sqrt{370} \approx 192.354 \)

\( v = \frac{90 \times 192.354}{190} \approx \frac{17311.86}{190} \approx 91.115 \approx 91.12 \)

So the speed is approximately 91.12 feet per second. Let's do it exactly:

\( y = \sqrt{190^2 + 30^2} = \sqrt{37000} = 10\sqrt{370} \)

\( v = \frac{90 \times 10\sqrt{370}}{190} = \frac{900\sqrt{370}}{190} = \frac{90\sqrt{370}}{19} \approx \frac{90 \times 19.235}{19} \approx \frac{1731.15}{19} \approx 91.11 \)

So the actual speed is approximately 91.1 feet per second (or more accurately, \( \frac{90\sqrt{370}}{19} \approx 91.1 \)).

Answer:

\boxed{91.1} (or more precise \( \frac{90\sqrt{370}}{19} \approx 91.1 \))