Question
A rocket having fuel as its bulk is initially at rest. Neglecting the effect of gravity, when fuel is burning at a constant rate, acceleration $a$ of the rocket with respect to time $t$ is best represented by one of the graphs given below.
Step 1
We have a rocket initially at rest, and its fuel is burning at a constant rate. We need to determine how the acceleration of the rocket changes with time, neglecting gravity. Show more…
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A rocket has total mass $M_{i}=360 \mathrm{~kg},$ including $M_{\text {fuel }}=$ $330 \mathrm{~kg}$ of fuel and oxidizer. In interstellar space, it starts from rest at the position $x=0$, turns on its engine at time $t=0,$ and puts out exhaust with relative speed $v_{e}=1500 \mathrm{~m} / \mathrm{s}$ at the constant rate $k=$ $2.50 \mathrm{~kg} / \mathrm{s} .$ The fuel will last for a burn time of $T_{b}=M_{\text {fuel }} / k=$ $330 \mathrm{~kg} /(2.5 \mathrm{~kg} / \mathrm{s})=132 \mathrm{~s}$. (a) Show that during the burn the velocity of the rocket as a function of time is given by $ v(t)=-v_{e} \ln \left(1-\frac{k t}{M}\right) $ (b) Make a graph of the velocity of the rocket as a function of time for times running from 0 to $132 \mathrm{~s}$. (c) Show that the acceleration of the rocket is $ a(t)=\frac{k v_{e}}{M_{i}-k t} $ (d) Graph the acceleration as a function of time. (e) Show that the position of the rocket is $ x(t)=v_{e}\left(\frac{M_{i}}{k}-t\right) \ln \left(1-\frac{k t}{M_{i}}\right)+v_{e} t $ (f) Graph the position during the burn as a function of time.
A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is $ m $, the fuel is consumed at rate $ r $, and the exhaust gases are ejected with constant velocity $ v_e $ (relative to the rocket). A model for the velocity of the rocket at time $ t $ is given by the equation $$ v(t) = -gt - v_e \ln \frac{m - rt}{m} $$ where $ g $ is the acceleration due to gravity and t is not too large. If $ g = 9.8 m/s^2 $, $ m = 30,000 kg $, $ r = 160 kg/s $, and $ v_e = 3000 m/s $, find the height of the rocket one minute after liftoff.
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Integration by Parts
A rocket has total mass $M_{i}=360 \mathrm{kg}$ , including $M_{f}=330 \mathrm{kg}$ of fuel and oxidizer. In interstellar space, it starts from rest at the position $x=0$ , turns on its engine at time $t=0$ , and puts out exhaust with relative speed $v_{e}=1500 \mathrm{m} / \mathrm{s}$ at the constant rate $k=2.50 \mathrm{kg} / \mathrm{s}$ . The fuel will last for a burn time of $T_{b}=M_{f} / k=330 \mathrm{kg} /(2.5 \mathrm{kg} / \mathrm{s})=132 \mathrm{s}$ . (a) Show that during the burn the velocity of the rocket as a function of time is given by $$ v(t)=-v_{e} \ln \left(1-\frac{k t}{M_{i}}\right) $$ (b) Make a graph of the velocity of the rocket as a function of time for times running from 0 to 132 s. (c) Show that the acceleration of the rocket is $$ a(t)=\frac{k v_{e}}{M_{i}-k t} $$ (d) Graph the acceleration as a function of time. (e) Show that the position of the rocket is $$ x(t)=v_{e}\left(\frac{M_{i}}{k}-t\right) \ln \left(1-\frac{k t}{M_{i}}\right)+v_{e} t $$ (f) Graph the position during the burn as a function of time.
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