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Chapter 2: Problem 75

A jet touches down on a runway with a speed of \(142.4 \mathrm{mph} .\) After \(12.4 \mathrm{~s},\) the jet comes to a complete stop. Assuming constant acceleration of the jet, how far down the runway from where it touched down does the jet stand?

Short Answer

Expert verified
Answer: The jet stands about 787.44 meters down the runway after decelerating to a complete stop.

Step by step solution

01

Identify the known values

The problem provides the following information: - Initial velocity of the jet, \(v_i = 142.4\,\mathrm{mph}\) - Time of acceleration, \(t = 12.4\,\mathrm{s}\) - Final velocity, \(v_f = 0\) since the jet comes to a complete stop.
02

Convert the initial velocity to \(\mathrm{m/s}\)

Since we have the time in seconds, we need to convert the initial velocity from \(\mathrm{mph}\) to \(\mathrm{m/s}\): $$v_i = 142.4\,\mathrm{mph} \cdot 1609.34 \mathrm{m/mile} \cdot \dfrac{1\,\mathrm{h}}{3600\,\mathrm{s}} ≈ 63.6\,\mathrm{m/s}$$
03

Find the acceleration

Using the formula \(v_f = v_i + at\), we can solve for acceleration \(a\): $$a = \dfrac{v_f - v_i}{t} = \dfrac{0 - 63.6\,\mathrm{m/s}}{12.4\,\mathrm{s}} ≈ -5.14\,\mathrm{m/s^2}$$
04

Find the distance traveled

Finally, use the equation \(d = v_i t + 0.5at^2\) to find the distance the jet traveled on the runway: $$d = 63.6\,\mathrm{m/s}\cdot 12.4\,\mathrm{s} + 0.5\cdot (-5.14\,\mathrm{m/s^2})(12.4\,\mathrm{s})^2 ≈ 787.44\,\mathrm{m}$$ So, the jet stands about \(787.44\,\mathrm{m}\) down from its touchdown on the runway.

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