4. (5 points) Rocket Motion A rocket, fired upward from rest at time t 0, as an

Question

4. (5 points) Rocket Motion A rocket, fired upward from rest at time t 0, as an initial mass of m (including its fuel). Suppose we make the following assumptions: The fuel is consumed at a constant rate k 0. Air resistance is neglected. The fuel gases are expelled at a constant speed c relative to the rocket (a) The equation of motion for a rocket whose mass, m varies with time, t is given by: dm m(t)a. dt net where vr is the velocity of the exhaust fuel gases relative to the rocket, and Fnet is the net external force on the rocket. Choose the upward direction to be positive and write a first-order differential equation that models the velocity of the rocket. Use a dv/dt

Explanation / Answer

a>

the mass of the rocket includes the mass of the fues al well , and its given as = m(t)

as time passes by the mass m(t) decreases as the fuel is being consumes , this tell us that the mass is a function of time.

we are given that the fuel is being consumed at a constant rate = k , and k >0

=> dm(t)/dt = k

at t = 0 , m(t=0) = mo

and vr = c {given}

as the rocket goes up so gravity force will act on it in the downward direction = mg = (mass of the rocket)*g = mg

since there is no other force acting as the air resistance is =0

so Fnet = mg as forces need to be balanced

we are give that a = dv/dt

=> m(t)*a =vrdm/dt + Fnet

m(t)dv/dt = c*(k) + mg

dv/dt = kc/m + g     --------> first order differential equation that models the velocity of the rocket

b>

dv/dt = kc/m + g

dv = [kc/m + g]dt

integrating both sides

v = kcln[m(t)] + gt + C

we are give that at t=0 , m = mo and v = 0

=> 0 = kcln[mo] - 0 +C

=> C = - kcln[mo]

so the integral becomes

v = kcln[m(t)] + gt - kcln[mo]

v = kc{ln[m(t)] - kcln[mo] + gt

v = kc*ln[m(t)/mo] + gt

c>

the fuel accounts for 75% of the initial mass m(t) = .75mo

k = dm(t)/dt = 3/4*1/90 = 0.0083

c = 2500 m/s

g = 9.8 m/s^2

t=90 sec

v = kc*ln[m(t)/mo] + gt

v = .0083*2500*ln[3mo/4mo)] + 9.8*90

v = .0083*2500*ln[3/4] + 9.8*90

v = 876.030596 m/sec

or v = 876 m/sec

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