A rocket has total mass Mi = 360 kg, including Mfuel = 330 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 ve = 1 500 m/s at the constant rate k = 2.50 kg/s. The fuel will last for a burn time of Tb = Mfuel/k = 330 kg/(2.5 kg/s) = 132 s. (a) Show that during the burn the velocity of the rocket as a function of time is given by
(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
(d) Graph the acceleration as a function of time. (c) Show that the position of the rocket is
(f) Graph the position during the burn as a function of time.
(a)
The velocity of the rocket is
Answer to Problem 9.64P
The velocity of the rocket is
Explanation of Solution
The total mass of rocket is
Rocket works on Newton’s third law, the thrust generate by the exhaust creates a reaction which causing the rocket to fly.
The basic expression of the rocket propulsion is,
Here,
Formula to calculate the total mass of rocket is,
Here,
Conclusion:
Substitute
Therefore, The velocity of the rocket is
(b)
The graph of the velocity of the rocket as a function of time for times running from 0 to
Answer to Problem 9.64P
The graph of the velocity of the rocket as a function of time for times running from 0 to
Explanation of Solution
The velocity of the rocket is,
Substitute
The value of
0 | 0 |
10 | 107.95 |
20 | 224.28 |
30 | 350.39 |
40 | 488.09 |
50 | 639.72 |
60 | 808.43 |
70 | 998.53 |
80 | 1216.27 |
90 | 1471.08 |
100 | 1778.21 |
110 | 2164.86 |
120 | 2687.16 |
130 | 3495.24 |
132 | 3726.3 |
Draw the graph between
Figure I
Conclusion:
Therefore, the graph of the velocity of the rocket as a function of time for times running from 0 to
(c)
The acceleration of the rocket is
Answer to Problem 9.64P
The acceleration of the rocket is
Explanation of Solution
The velocity of the rocket is,
Formula to calculate the acceleration is,
Conclusion:
Substitute
Therefore, the acceleration of the rocket is
(d)
The graph of the acceleration of the rocket as a function of time for times running from 0 to
Answer to Problem 9.64P
The graph of the acceleration of the rocket as a function of time for times running from 0 to
Explanation of Solution
The acceleration of the rocket is,
Substitute
The value of
0 | 10.42 |
10 | 11.12 |
20 | 12.09 |
30 | 13.15 |
40 | 14.42 |
50 | 15.95 |
60 | 17.85 |
70 | 20.27 |
80 | 23.43 |
90 | 27.78 |
100 | 34.09 |
110 | 44.12 |
120 | 62.5 |
130 | 107.14 |
132 | 125 |
Draw the graph between
Figure II
Conclusion:
Therefore, the graph of the acceleration of the rocket as a function of time for times running from 0 to
(e)
The position of the rocket is
Answer to Problem 9.64P
The position of the rocket is
Explanation of Solution
The velocity of the rocket is,
Formula to calculate the position is,
Substitute
Conclusion:
Therefore, the position of the rocket is
(f)
The graph of the position of the rocket as a function of time for times running from 0 to
Answer to Problem 9.64P
The graph of the position of the rocket as a function of time for times running from 0 to
Explanation of Solution
Rocket works on Newton’s third law the thrust generate by the exhaust creates a reaction which causing the rocket to fly.
The position of the rocket is,
Substitute
The value of
0 | 0 |
10 | 534.2866 |
20 | 2189.017 |
30 | 5054.74 |
40 | 9237.945 |
50 | 14865.69 |
60 | 22092.2 |
70 | 31108.67 |
80 | 42158.38 |
90 | 55561.47 |
100 | 71758.44 |
110 | 91394.47 |
120 | 115508.2 |
130 | 146066.6 |
132 | 153284.3 |
Draw the graph between
Figure III
Conclusion:
Therefore, the graph of the position of the rocket as a function of time for times running from 0 to
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Chapter 9 Solutions
EBK PHYSICS:F/SCI.+ENGRS.,TECH.UPDATED
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