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Vika [28.1K]
3 years ago
11

A jumbo jet must reach a speed of 360 km/h (= 225 mi/h) on the runway for takeoff. assuming a constant acceleration and a runway

1.50 km long, what minimum acceleration from rest is required?
Physics
1 answer:
Alinara [238K]3 years ago
3 0
We just consider the moves of the jumbo jet, not it's causes (we solve it kinematically)

It has a constant acceleration, so then we may use this equation to solve our problem: a = Vf^2 - Vo^2 / 2∆x

Vf (The max velocity that allowed) = 360 Km/h
Vo = 0 (At rest)
∆x = 1.5 Km

then..

a = (360)^2 / (2)(1.5) = 43200 Km/h^2

Convert those numbers to m/sec^2 you will surprised by the results=))
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Solution :

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According to the conservation of momentum,

Mv+mu=Mv_1+mv_2

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\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2    ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$  

21681 = 9v_1^2+2v_2^2

Substituting (i) in (ii)

21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2

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(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0

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Solving the equation, we get

v_2=96 \ m/s, -30 \ m/s

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Therefore, substituting 96 m/s for v_2 in (i), we get

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