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 the fuel) is m, the fuel is consumed at rate r, and the exhaust gasses are ejected with constant velocity v, (relative to the rocket) A model for the velocity of the rocket at time t is given by v(t)=-gt-v ln(m-rt/m) where g is the acceleration due to gravity and t is not to large. If g=9.8m/s^2 m=28,000 kg, r=190 kg/s, and v=3,100 m/s. Find the height of the rocket one minute after lift off. (round answer to nearest whole meter)

Answer 1

You can use energy or kinetics?

Answer 2

Its a Calc 2 class I don't really know either...

Answer 3

It looks like you just have to integrate.

Answer 4

Consider it's height at time \(0\) to be \(0\)

Answer 5

I thought I had to integrate he vt equation then put in the given variables then integrate again but I was wrong

Answer 6

\[
h=h(60)-h(0)=\int_0^{60} v(t)\;dt
\]

Answer 7

Shall i try?

Answer 8

ya its solved

Answer 9

@cydneehill Did you integrate correctly?

Answer 10

I checked it twice and I thought so

Answer 11

v(t) = -gt - ve(ln((m - rt)/m))

Then:
x(t) = ∫ v(t) dt
∫ [-gt - ve(ln((m - rt)/m))] dt

Let u = (m - rt)/m
du = (-r/m) dt
dt = (du)/(-r/m) = (-m/r) du

x(t) = (-1/2)gt^2 - (ve(-m/r)(u(lnu - 1) + C
= ((mve)/r)((m - rt)/m)(ln((m - rt)/m) - 1) - (1/2)gt^2
= (ve(m - rt)/r) (ln((m - rt)/m) - 1) - (1/2)gt^2

g = 9.8 m/s^2
m = 27000 kg
r = 130 kg/s
ve = 2700 m/s
t = 60 s

x(60) = (2700(27000 - (130*60))/130) ((ln((27000 - (130*60))/27000)) - 1) - ((1/2)(9.8)(60^2))
= (((2700(27000 - 7800)/130) (ln((27000 - 7800)/27000) - 1)) - (16560)
= ((2700(19200))/130)) (ln(19200/27000) - 1) - (16560)
= ((398769.231)(0.389446386)) - (16560)
= 138739.235860949
= 138739 m

It said to round to the nearest whole meter. In scientific notation with significant figures, the answer is: 14 x 10^4 km

Answer 12

\[\int\limits_{0}^{60}-g t-v \ln(\frac{ m-rt }{ m }) = -60 (g t+v \ln(1-\frac{ rt }{ m })\]

Answer 13

@cydneehill I don't think you're using the right anti derivative.

Answer 14

Integrate \(-gt\) and \(-v\ln\frac{m-rt}{m}\) separately

Answer 15

from 0 to 60?

Answer 16

Yeah.

Answer 17

first one would be -1800g the second one \[\frac{ v((m-60r)\ln(1-\frac{ 60r }{ m }) +60r}{ r }\]

Answer 18

oh \[\int\limits_{0}^{60} -g t- v \ln \frac{ m-rt }{ m }= \frac{ 60r(v-30g)+v(m-60r)\ln(1-\frac{ 60r }{ m }) }{ r }\]?

Answer 19

so do I plug my given into that equation and integrate again? or just plug it in and get my answer

Answer 20

plug in my given variables and get my answer thanks