Chapter 7.1, Problem 68E

### Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

Chapter
Section

### Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

# 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 ) = − g t − v e   ln m − r t 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.

To determine

To find: the height of the rocket after one minute of takeoff.

Explanation

Recall that velocity is the rate at which the position of an object changes. It is given by the ratio of distance traveled to the time taken in covering that distance. Then, the distance covered will be the product of time and velocity. Since the rocket is moving in vertical direction, the product of time and rocket’s velocity will give the height of the rocket at that particular time.

Formula used:

If v(t) is the velocity which changes with time, then distance traveled in time t will be given by the following integral

d=0tv(t)dt

Given:

Velocity of the rocket at time t, v(t)=gtvelnmrtm

Acceleration due to gravity and mass, g=9.8 m/s2,m=30000 kg

Fuel consumption rate and exhaust gas velocity, r=160 kg/s, ve=3000 m/s

Calculation:

Substitute the expression for velocity v in the formula for distance

d=0T(gtvelnmrtm)dt=g0Ttdtve0Tln(mrtm)dt

The height of the rocket one minute after liftoff is required, so substitute T as 60 (1 min=60 s)

d=g060tdtve060ln(mrtm)dt

Let p=mrtm, then its differentiation will be dp=rdtm. Then, the integral for distance will become:

d=g060tdt+ve060lnpmrdp=g060tdt+vemr060lnpdp …… (1)

Solve the integral in the last term using integration by parts. The technique of integration by parts comes in handy when the integrand involves product of two functions. It can be thought of as a rule corresponding to the product rule in differentiation. The formula for integration by parts for definite integral is given by

abf(x)g(x)dx=f(x)g(x)]ababg(x)f(x)dx

Make the choice for u and dv such that the resulting integration from the formula above is easier to integrate

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