Answer to Question #136659 in Mechanics | Relativity for abdulwahab

Question

The motion of a rocket burning its fuel at a constant rate while moving through empty
interstellar space can be described by
x(t) = uext + uex
1
b
− t

ln(1 − bt).
uex and b are constants (uex is the exhaust velocity of the gases at the tail of the rocket,
and b is proportional to the rate of fuel consumption). For this question, you should
know that ln is the inverse of the exponential function, that is, ln(e
x
) = x. Also,
d
dx ln x =
1
x
. [10]
a) What happens when t > 1/b? Explain.
b) Find a formula for the instantaneous velocity of the rocket.
c) Find a formula for the instantaneous acceleration.
d) Suppose that a rocket with uex = 3.0 × 103 m/s and b = 7.5 × 10−3
s
−1
takes 120
s to burn all its fuel. What is the instantaneous velocity at t = 0 s? At t = 120 s?

Answer 1

As per the given question,

$x(t)=u_{ex}\left(\dfrac{1}{b}-t\right)\ln\left(1-bt\right)+u_{ex}t$

a) As per the question, if $t>\frac{1}{b}$

$x(t)=u_{ex}\left(\dfrac{1}{b}-t\right)\ln\left(1-bt\right)+u_{ex}t$

so tb>1 hence first term will not defined because ln is not defined for negative value.

b)

Now taking the differentiation with respect to t

$\frac{x(t)}{dt}=\frac{d }{dx}[(u_{ex}\left(\frac{1}{b}-t\right)\ln\left(1-bt\right)+u_{ex}t)]$

$\Rightarrow$ $dv=u_{ex}\frac{d}{dt}[(\left(\frac{1}{b}-t\right)\ln\left(1-bt\right)]+u_{ex}$

$\Rightarrow$ $dv=u_{ex}[\frac{d}{dt}[(\frac{1}{b}-t)\ln\left(1-bt\right)+[(\frac{1}{b}-t)\frac{d}{dt}\ln(1-bt)]+u_{ex}$

$\Rightarrow dv= u_{ex}\left(-\ln\left(1-bt\right)-\dfrac{b\left(\frac{1}{b}-t\right)}{1-bt}\right)+t$

$\Rightarrow dv=-u_{ex}\ln\left(1-bt\right)-\dfrac{bu_{ex}\left(\frac{1}{b}-t\right)}{1-bt}+u$

$\Rightarrow dv= -u\ln\left(1-bt\right)$

c) For the instantaneous acceleration,

$da=\frac{dv}{dt}$

$\Rightarrow da=\frac{d}{dt} \left( -u_{ex} \ln \left(1-bt\right) \right)$

$\Rightarrow da=-u_{ex}\frac{d}{dt}\left( \ln(1-bt)\right)$

$\Rightarrow da=-u_{ex}(\frac{1}{1-bt})\frac{d}{dt}(1-bt)$

$\Rightarrow da=\dfrac{bu_{ex}}{1-bt}$

d) Now it is given

$u_{ex} = 3.0 × 10^3 m/s$

$b = 7.5 × 10^{−3}/sec$

$t=0$

$t=120 sec$

$\Rightarrow dv= -u\ln\left(1-bt\right)$

at t=0

$dv=3.0\times 10^3\ln(1-7.5\times 10^{-3}\times 0)$

$\Rightarrow dv=3.0\times 10^3\ln(1)$

$\Rightarrow dv=0$

Now,

at t=120

$dv=3.0\times 10^3\ln(1-7.5\times 10^{-3}\times 120)$

$\Rightarrow dv=3.0\times 10^3\ln(1-0.016)$

$\Rightarrow dv=0.016\times 3\times 10^3 m/sec$

$\Rightarrow dv=-48m/sec$