Question

A 2400-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 1.90 m/s2 and 1.20 s later an upward velocity of 1.50 m/s .

Part A) Determine A.

Part B) Determine B.

Part C) At 3.30 s after fuel ignition, what is the acceleration of the rocket?

Part D) At 3.30 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?

Answer 1

Solution:

A) Given at the instant of ignition a = 1.90 m/s^2

=> dv / dt = 1.90

=> d( At + Bt^2 ) /dt = 1.90

=> A + 2Bt = 1.90

=> at ignition t = 0 so,

=> A = 1.90

b) at t = 1.20 s

=> v(t) = At + Bt^2

=> 1.50 = 1.90 * 1.20 + B ( 1.20)^2

=> B = -0.542

C) a = A + 2Bt

=> a = 1.90 - 2 * 0.542 * 3.3

=> a = -1.6772 m/s^2

D) F = m(g-a)

=> F = 2400 ( 9.8 + 1.6772 )

=> F = 27545.28 N

E) weight of the rocket = 2400 * 9.8

=> W = 23520 N

F = 1.711 W

F) F = m(g-a)

=> F = m (9.8 + 1.90)

=> F = 2400 ( 11.7)

=> F = 28080 N