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Topic: seaplane question (Read 625 times) |
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nikki
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A sea plane lands on water with a touch down velocity of 160 km/hr and slow to a velocity of 26km/hr in a distance of 380m. If the deceleration of the plane in the water is proportion to Kv^2, where K is a constant, find how long it takes to travel the 380m
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Sir Col
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Re: seaplane question
« Reply #1 on: May 25th, 2004, 10:41am » |
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As a = dv/dt = kv2, begin by separating the variables and integrating.
[int] 1/v2 dv = [int] k dt
-1/v = kt + c
When t = 0, v = 160 km/h, so c = -1/160. Rearranging we get v = 160/(1-kt)
When v = 26 km/h (and s = 380 m), 26 = 160/(1-kt), so 1-kt = 160/26 = 80/13.
As v = ds/dt = 160/(1-kt), [int] ds = [int] 160/(1-kt) dt, s = -(160/k)ln(1-kt).
When s = 380 m = 0.38 km, 1-kt = 80/13, so, 0.38 = -(160/k)ln(80/13).
Therefore, k = -(160/0.38 )ln(80/13) = -(8000/19)ln(80/13) [approx] -765.
Using 26 = 160/(1+765t), we get t [approx] 0.00674 hours [approx] 24 seconds.
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Nikki
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thanku sOOOOoo muCH! 
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