A 2570-kg rocket-propelled capsule moves in the vertical direction as it launche

Question

A 2570-kg rocket-propelled capsule moves in the vertical direction as it launches from the space center. Its fuel (of negligible mass) provides a thrust force so that its velocity in the vertical direction (as a function of time) is given by v(t)=At+Bt2, where A and B are constants and t  is the time elapsed since the fuel is ignited. At the instant of ignition, the rocket has an upward acceleration of 1.50m/s2 and 2.00s later an upward velocity of 2.48m/s .

Part A) Determin A.

Part B) Determine B.

Part C) At 4.50s after fuel ignition, what is the acceleration of the rocket?

Part D) At 4.50s after fuel ignition, what thrust force does the burning fuel exert on it? Assume no air resistance. Express the thrust in newtons.

Part E) What thrust force does the burning fuel exert on it? Assume no air resistance. Express the thrus as a multiple of the rocket's weight.

Part F) What was the initial thrust due to the fuel?

Explanation / Answer

a(t) = dV/dt

a = A + 2Bt

at t =0 a = 1.5

1.5 = A +0

A = 1.5 m/s^2

------------partB)

t= 2s

v = 2.48

2.48 = 1.5*2 + B*2*2 = 3 + 4B

B = 2.48 - 3/4 = -0.13m/s^3

-------------

C)a = 1.5+(2*-0.13*4.5) = 2.915 m/s^2

D) T = m*a = 2570*2.915 = 7491.55 N

E) T = ma*mg/mg =   7491.55w/(2570*9.8) = 0.29w

F) T = 2570*1.5 = 3855 N

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