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I'm not even sure where to begin.

To help study damage to an aircraft that collide with birds you design a test gun that will accelertae chicken sized objects so that their displacement along the gun barrel is given by
X=(9.0*10^3m/s^2)t^2-(8.0*10^4m/s^3)t^3
The object leaves the end of the barrel at t=0.03 s.

a) How long must the gun barrel be.
b) What will be the speed of the object as they leave the end of the barrel.
c)What net force must be exerted on a 1.5kg object at t=0.03s.

2006-11-08 14:43:48 · 2 answers · asked by lpfanz89 1 in Science & Mathematics Physics

2 answers

distance

X=(9.0*10^3m/s^2)t^2-(8.0*10^4 m/s^3)t^3, plug 0.03 s
=0.0009(9000) - 0.000027(80000)
= 8.1 - 2.16
=5.94 m, length = 5.94 m

velocity

dx/dt=18000t - 270000t^2, plug in 0.03 s
= 297 m/s

acceleration

d2x/dt2 = 18000 - 540000t, plug in 0.03 s

a=1800 m/s2

so at t=0.03 s, F = mxa = 1.5 * 1800 = 2700 N

2006-11-09 08:05:57 · answer #1 · answered by Mech_Eng 3 · 0 0

a) you already have the equation for distance
b) Derivative of distance is velocity
c) Derivative of speed is acceleration

2006-11-08 14:58:09 · answer #2 · answered by arbiter007 6 · 0 0

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