A 2190-kg test rocket is launched vertically from the launch pad. Its fuel (of n

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A 2190-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 1.20 m/s2 and 1.50s later an upward velocity of 2.06m/s . Determine A. Submit My Answers give Up Determine B. Submit My Answers give Up At 4.005 after fuel ignition, what is the acceleration of the rocket? Submit My Answers give Up A 2190-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 1.20 m/s2 and 1.50s later an upward velocity of 2.06m/s . What thrust force does the burning fuel exert on it, assume no air resistance? Express the thrust in newtons. Submit My Answers Give UP 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. Submit My Answers Give UP What was the initial thrust due to the fuel? Submit My Answers Give UP

Explanation / Answer

v(t) = A*t + B*t^2 debba(v)/debba(t) = a(t) = A + 2*B*t putting the given conditions ,we get B = 0.11 m/sec^3 C) Putting the t = 4.00 sec in the equation a(t) = A + 2*B*t , we get at 4.00 seconds, a = 2.08 m/sec^2 D) the thrust on the rocket = m*(a+g) at an instant of ignition, thrust = 2190*(1.20+9.8) =24090 N t t= 4 seconds, thrust = 2190*(2.08+9.8) =26017 N E) in terms of weight, thrust on the rocket at an instant of ignition = 1.122*W thrust on the rocket at t=4 sec = 1.21*W F) Initial thrust on the rocket = 24090 N

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