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3 years ago

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A jet accelerates at a=3.92 m/s2 from rest until it reaches its takeoff velocity of 360 km/hr. It has to travel for 5 seconds at

its takeoff velocity before taking off. What's the minimum distance that the jet will travel on the runway?

1 answer:

Art [367]3 years ago

7 0

Answer:

$\Delta s = 1775.510\,m$

Explanation:

The minimum distance for takeoff is:

$\Delta s = \Delta s_{1} + \Delta s_{2}$

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a} + v_{f}\cdot \Delta t$

$\Delta s = \frac{(100\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}}{2\cdot (3.92\,\frac{m}{s^{2}})}+(100\,\frac{m}{s} )\cdot (5\,s)$

$\Delta s = 1775.510\,m$

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${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

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${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

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${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

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$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

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$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

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$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

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$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

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$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

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$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

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$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

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$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

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$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

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$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

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