Concept explainers
Rocket Motion—Continued In Problem 39 suppose of the rocket’s initial mass m0 that 50 kg is the mass of the fuel.
- (a) What is the burnout time tb, or the time at which all the fuel is consumed?
- (b) What is the velocity of the rocket at burnout?
- (c) What is the height of the rocket at burnout?
- (d) Why would you expect the rocket to attain an altitude higher than the number in part (b)?
- (e) After burnout what is a mathematical model for the velocity of the rocket?
39. Rocket Motion Suppose a small single-stage rocket of total mass m(t) is launched vertically, the positive direction is upward, the air resistance is linear, and the rocket consumes its fuel at a constant rate. In Problem 22 of Exercises 1.3 you were asked to use Newton’s second law of motion in the form given in (17) of that exercise set to show that a mathematical model for the velocity v(t) of the rocket is given by
where k is the air resistance constant of proportionality, λ is the constant rate at which fuel is consumed, R is the thrust of the rocket, m(t) = m0 − λt, m0 is the total mass of the rocket at t = 0, and g is the acceleration due to gravity.
- (a) Find the velocity v(t) of the rocket if m0 = 200 kg, R = 2000 N, λ = 1 kg/s, g = 9.8 m/s2, k = 3 kg/s, and v(0) = 0.
- (b) Use ds/dt = v and the result in part (a) to find the height s(t) of the rocket at time t.
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Chapter 3 Solutions
Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage