As of 2012, all of the shuttles in NASA\'s space shuttle program have been retir

Question

As of 2012, all of the shuttles in NASA's space shuttle program have been retired. When they once launched, a space shuttle (mass = 2.03 times 10^6 kg) would lift off as the thrust of its three main engines was suddenly augmented by the firing of the solid rocket boosters at t = 0 s. Suppose the magnitude of the shuttle's acceleration is given by a = 3.10t + 0.315t^2 - 0.130t^3 where a is in meters per second squared and t is in seconds, in the first few seconds after liftoff. What is the change in the kinetic energy of the space shuttle during the first 2.45 s after liftoff in this scenario?

Explanation / Answer

accelaration is a = (3.1*t) + (0.315*t^2) - (0.130*t^3)

velocity of rocket is v = integral of (a*dt)

v = (3.1*t^2/2)+(0.315*t^3/3) - (0.13*t^4/4)

at t = 2.45 sec

v = (3.1*2.45^2/2)+(0.315*2.45^3/3) - (0.13*2.45^4/4)

v = 9.67 m/s is the speed at t = 2.45 sec

at t = 0 sec ,initial speed is Vo = 0 m/sec

change in kinetic energy is dK = Kf-Ki

Kf is the final kinetic energy = 0.5*m*V^2 = 0.5*2.03*10^6*9.67^2 = 9.49*10^7 J

Ki is the initial kinetic energy = 0.5*m*Vo^2 = 0.5*2.03*10^6*0^2 = 0 J

then change in kinetic energy is dK = 9.49*10^7 J

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