A 2590-kg test rocket is launched vertically from the launch pad. Its fuel (of n

Question

A 2590-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 + Bt^2, 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 1.70 m/s^2 and 1.00 s later an upward velocity of 1.80 m/s. Determine A and B, including their SI units. At 4.00 s after fuel ignition, what is the acceleration of the rocket? m/s^2 What thrust force does the burning fuel exert on it, assume no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. What was the initial thrust due to the fuel?

Explanation / Answer

Given,

Mass of the rocket = M = 2590 kg

Velocity of the rocket is given by

v(t) = At + Bt²

Therefore, the acceleration of the rocket is given by

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

(a) Also,

At the instant of ignition, the rocket has an upward acceleration of 1.70 m/s2

==> a(0) = A + 2(B)(0) = 1.70 m/s²

==> A = 1.70 m/s2

Given,

1 second later an upward velocity of 1.8 m/s

==> v(1) = (1.70 m/s²)(1 s) + B*(1 s)² = 1.8 m/s

==> B = 0.1 m/s3

Therfore,

v(t) = 1.7t + 0.1t2

a(t) = 1.7 + 0.2t

(b) Now, after 4 seconds of rocket ignition, the acceleration is given by

a(4) = 1.7 + 0.2*(4) = 2.5 m/s2

(c) Now, 4 seconds after fuel ignition, The thrust force the burning fuel exert on the rocket is given by

ma = T - mg

(2590 kg)*(2.5m/s²) = T - (2590 kg)*(9.81 m/s²)

T = 31882.9 N

Now, Thrust in terms of rocket weight is given by

==> T/(mg) = 31882.9 N / [(2590 kg)*(9.81 m/s2)]

= 1.254841998 (factor of the rocket's weight)

(d) The initial thrust due to fuel is given by

ma = T - mg

(2590 kg)*(1.7 m/s²) = T - (2590 kg)(9.81 m/s²)

==> T = 29810.9 N

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