A 2810-kg test rocket is launched vertically from the launch pad. Its fuel (of n

Question

A 2810-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t)=At+Bt2, where A and B are constants and time is measured from the instant the fuel is ignited. At the instant of ignition, the rocket has an upward acceleration of 1.80 m/s2 and 1.90 s later an upward velocity of 2.35 m/s .

Part A: Determine A. --> I got this correct: 1.80 m/s2

Part B: Determine B. --> I got this correct: -0.296 m/s3

Part C: At 3.10 s after fuel ignition, what is the acceleration of the rocket? --> I keep getting this wrong on the online HW and I can't see why. My answer is:

at t = 3.10s, a=1.80 m/s2+2(-0.296 m/s3)(3.10 s) = -0.0352

Please help!

Also, for the rest of the problem...

Part D: At 3.10 s 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 thrust as a multiple of the rocket's weight.

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

Thanks in advance to all you respond!

Explanation / Answer

The velocity is given by:

v(t) = At + Bt2

=> a(t) = dv/dt = A + 2Bt

at t = 0s, a(t) = 1.8 m/s2

therefore, 1.8 = A + 2B(0)

=> A = 1.8 m/s2

so, v(t) = 1.8t + Bt2

and at t = 1.9s, v(t) = 2.35 m/s

=> 2.35 = 1.8(1.9) + B(1.9)2

=> B = - 0.2964 m/s3

therefore, v(t) = 1.8t - 0.2964t2

C] at, t = 3.1s,

a(t) = 1.8 + 2(-0.2964)(3.1) = - 0.0377 m/s2

D] F = ma = 2810 x -0.0377 = - 105.86 N

this is the thrust exerted by the burning fuel.

E] Rocket's weight = W = mg = - 2810 (9.8) = - 27538 N

therefore, the thrust exerted by burning fuel in terms of the weight will be: F = 105.86/27538 = 0.003844W.

F] Initial thrust due to the fuel will be,

F = 2810 (1.8) = 5058 N.

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