Nearly Impossible Calc Question? (I tried it, its true)?

Question

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 ve [v followed by little e] (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation below where g is the acceleration due to gravity and t is not too large.

v(t) = -gt - ve ln((m-rt)/m) [that's v then little e]

If g = 9.8 m/s2, m = 30000 kg, r = 160 kg/s, and ve = 3000 m/s, find the height of the rocket one minute after liftoff.

Ok, any ideas on where to start...? Dont you just plug in?

Answer 1

You have to find the height/position function. So integrate the velocity function with respect to time.

int[-gt - v[e] ln((m-rt)/m) dt]

= int[-gt dt] - int[v[e] ln((m-rt)/m) dt]
= int[-gt dt] - v[e] int[ln(1 - rt/m) dt]

int[-gt dt]
= (-gt^2) / 2

For the integral on the right:
let u = ln(1 - rt/m) dv = dt
du = r / (m - rt) dt, v = t

t ln(1 - rt/m) - int [rt / (m-rt) dt]

Now let w = m - rt
dw = -r dt
dt = -dw / r

= t ln(1 - rt/m) + (1/r) int [(m - w) / (w) dw]
= t ln(1 - rt/m) + (1/r) int [m/w - 1 dw]
= t ln(1 - rt/m) + (1/r) [m ln(w) - w]
= t ln(1 - rt/m) + (1/r) [m ln(m - rt) - m + rt]

So the complete integral is:
s(t) = (-gt^2)/2 + v[e] {t ln(1 - rt/m) + (1/r) [m ln(m - rt) - m + rt] + C}

If s(0) = 0
0 = (m/r) ln(m) - m/r + C
So C = -m/r ln(m) + m/r

Thus
s(t) = (-gt^2)/2 + v[e] {t ln(1 - rt/m) + (1/r) [m ln(m - rt) - m + rt] - m/r ln(m) + m/r}
= (-gt^2)/2 + v[e]{ t ln(1 - rt/m) + (m/r) ln(m - rt) - m/r + t - m/r ln(m) + m/r}
= (-gt^2)/2 + v[e] {t ln(1 - rt/m) + (m/r) ln(m - rt) + t - (m/r) ln(m)}

1 minute = 60 seconds
So plug t=60 into s(t) and that's your answer.

Answer 2

This doesn't look too complicated. It's a related rates problem, and your related rates (fuel consumption and mass) are already included in the calculation as the natural log of the mass of the rocket less the fuel consumption to the mass of the rocket.

You have everything you need to solve, just go for it. The only thing is, make sure you know in what units your TIME is denoted, so you know how many of them to count to make one minute.

Answer 3

plugging in gives you the velocity at time 60. you want the distance traveled, which is the integral as t goes from 0 to 60 of the expression.

Answer 4

h(t) = -1/2gt^2-ve Int(0,60 )ln(1-rt/m)dt
Int ln(1-r/m*t) = t*ln(1-r/m*t) +Int r/m*t/(1-r/mt) dt =-t-ln(1-r/mt)(m/r)
(0,60)
You can make the numerical substitutions