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Chapter 4: Problem 168

\(\mathrm{A}\) -home-made" solid propellant rocket has an initial mass of \(20 \mathrm{lbm} ; 15 \mathrm{lbm}\) of this is fuel. The rocket is directed vertically upward from rest, burns fuel at a constant rate of \(0.5 \mathrm{lbm} / \mathrm{s},\) and ejects exhaust gas at a speed of \(6500 \mathrm{ft} / \mathrm{s}\) relative to the rocket. Assume that the pressure at the exit is atmospheric and that air resistance may be neglected. Calculate the rocket speed after 20 s and the distance traveled by the rocket in 20 s. Plot the rocket speed and the distance traveled as functions of time.

Short Answer

Expert verified
After 20 seconds, the rocket will be moving at a speed of 325 ft/s and it will have traveled a distance of 6500 ft.

Step by step solution

01

Determine the mass of the rocket at any given time

Since the rocket burns fuel at a constant rate of 0.5 lbm/s, the mass of the remaining rocket (fuel plus rocket structure) after a time \(t\) is \(m(t) = 20 - 0.5t\) lbm. After 20 seconds, the mass of the rocket is \(m(20) = 20 - 0.5 * 20 = 10 lbm.\)
02

Determine the rate of change of the rocket's momentum

The momentum of the exhaust gas being expelled each second is given by the product of its mass and its velocity relative to the rocket (6500 ft/s). So, the rate of change of the rocket's momentum is \(dm*v = 0.5 * 6500 = 3250 lbm*ft/s^2.\)
03

Calculate the rocket's velocity after 20 seconds

By conservation of momentum, the magnitude of the rocket's velocity at any time \(t\) can be obtained by dividing the momentum transferred to it by the mass remaining at that time. So, the speed of the rocket after 20 seconds is \(v(20) = dm*v / m(20) = 3250 / 10 = 325 ft/s.\)
04

Calculate the distance traveled by the rocket in 20 seconds

The distance \(s(t)\) traveled by the rocket in the first \(t\) seconds is obtained by integrating the velocity with respect to time from 0 to \(t\). However, since the velocity derived is an average constant velocity for the 20 second span, we can use the formula for distance: \(s = vt\). That gives us \(s(20) = v(20) * 20 = 325 * 20 = 6500 ft\).

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