A 50-kg model rocket lifts off by expelling fuel downward at a rate of k = 4.75
Question:
A 50-kg model rocket lifts off by expelling fuel downward at a rate of k = 4.75 kg/s for 10 s. The fuel leaves the end of the rocket with an exhaust velocity of b = −100 m/s. Let m(t) be the mass of the rocket at time t. From the law of conservation of momentum, we find the following differential equation for the rocket’s velocity v(t) (in meters per second):
(a) Show that m(t) = 50 − 4.75t kg.
(b) Solve for v(t) and compute the rocket’s velocity at rocket burnout (after 10 s).
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a For 0 t 10 the rocket is expelling fuel at a constant rate of 475 kgs giving m t 475 Hence mt ...View the full answer
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