Question

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a person is standing 400 feet away from a model rocket that is fired straight up into the air at a rate of 12 ft/sec. at what rate is the distance between the person and the rocket increasing 1 minute after liftoff?

a. 4.3
b. 10.5
c. 68.6
d. 78.5

Answer:

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<reasoning>

Set up the geometric model

We model the scenario using a right triangle.
Let \(x\) be the constant horizontal distance from the observer to the launch pad:
\[x = 400\text{ ft}\]
Let \(y(t)\) be the height of the rocket at time \(t\).
Let \(s(t)\) be the distance between the observer and the rocket at time \(t\).
By the Pythagorean theorem:
\[s^2 = x^2 + y^2\]

Determine values at the given instant

The rocket rises at a constant rate:
\[\frac{dy}{dt} = 12\text{ ft/sec}\]
At \(t = 1\text{ minute} = 60\text{ seconds}\), the height \(y\) is:
\[y = 12\text{ ft/sec} \times 60\text{ sec} = 720\text{ ft}\]
Using the Pythagorean theorem, find the distance \(s\) at this instant:
\[s = \sqrt{400^2 + 720^2} = \sqrt{160000 + 518400} = \sqrt{678400} \approx 823.65\text{ ft}\]

Differentiate with respect to time

Differentiate both sides of the relation \(s^2 = x^2 + y^2\) with respect to \(t\):
\[2s\frac{ds}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}\]
Since the observer is stationary, \(x\) is constant, so \(\frac{dx}{dt} = 0\):
\[s\frac{ds}{dt} = y\frac{dy}{dt}\]

Solve for the rate of change of distance

Isolate \(\frac{ds}{dt}\):
\[\frac{ds}{dt} = \frac{y}{s}\frac{dy}{dt}\]
Substitute the known values at \(t = 60\text{ seconds}\):
\[\frac{ds}{dt} = \frac{720}{823.65} \times 12 \approx 10.49\text{ ft/sec}\]

Match with the given options

Rounding to one decimal place:
\[\frac{ds}{dt} \approx 10.5\text{ ft/sec}\]
This matches option B.
</reasoning>

<answer>
<mcq-option>(A) 4.3</mcq-option>
<mcq-correct>(B) 10.5</mcq-correct>
<mcq-option>(C) 68.6</mcq-option>
<mcq-option>(D) 78.5</mcq-option>
</answer>

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