Question

A jet touches down on a runway with a speed of 142.4mph. after 12.4s, 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?

Answer 1

Answer on Question #70329, Physics / Other

A jet touches down on a runway with a speed of 142.4mph. after 12.4s, 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?

SOLUTION

Uniform acceleration is a type of motion in which the velocity of an object changes by an equal amount in every equal time period. There are simple formulas elating the displacement, initial and time-dependent velocities, and acceleration to the time elapsed:

$$\vec {v} (t) = \overrightarrow {v _ {0}} + \vec {a} t, (1)$$ $$v ^ {2} (t) = v _ {0} ^ {2} (t) + 2 \vec {a} (\vec {s} (t) - \overrightarrow {s _ {0}}), (2)$$

where $t$ is the elapsed time,

$\overrightarrow{s_0}$ is the initial displacement from the origin,

$\vec{s}(t)$ is the displacement from the origin at time $t$ ,

$\overrightarrow{v_0}$ is the initial velocity, and

$\vec{a}$ is the uniform rate of acceleration.

Let us present a picture to imagine the task.

Fig. 1. Jet lending

A jet touches down on a runway with a speed of $v_0 = 142.4 \, \text{mph} = (142.4 / 3600) \, \text{mile/s} \approx 0.04 \, \text{mile/s}$ . Let us suppose that it is a time $t = 0$ , and the touch point has a displacement $s_0 = 0$ . After $t_{\text{end}} = 12.4 \, \text{s}$ , the jet comes to a complete stop: $v_{\text{end}} = 0$ . Assuming constant acceleration of the jet, from the formula (1) we can find acceleration:

$$\vec {a} = \frac {\vec {v} (t _ {e n d}) - \vec {v _ {0}}}{t _ {e n d}}, (3)$$ $$a = \frac {\left(0 - 0 . 0 4 \frac {m i l e}{s}\right)}{1 2 . 4 s} = - 0. 0 0 3 2 \frac {m i l e}{s ^ {2}}.$$

Appearing the minus means that acceleration decreases velocity. Extracting the displacement from the formula (2), we obtain a result:

$$s - s _ {0} = \frac {v _ {e n d} ^ {2} - v _ {0} ^ {2}}{2 a} = - \frac {\left(0 . 0 4 \frac {m i l e}{s}\right) ^ {2}}{2 (- 0 . 0 0 3 2 \frac {m i l e}{s ^ {2}})} = 0. 2 5 m i l e.$$

ANSWER: the jet stands 0.25 mile from the touched down point.

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