Question

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?

Short answer

a) $t_b=50$seconds

b)$v_b=70\text{m}/\text{s}$

c)$s_b=10,500\text{m}$

d) Because of thrust.

e)$v\left(t\right)=-195e^{\frac{1}{75}t}+265$

Step-by-step solution

Definition of Series circuits

The linear differential equation for the current i(t),

$\mathit{L}\frac{\mathbf{d}\mathbf{i}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{R}\mathit{i}\mathbf{=}\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}$

where L and R are known as the inductance and the resistance, respectively. The current i(t) is also called the response of the system.

Differentiate the equation

a) Since we know that from the differential equation showed in the statement of problem (39), we have the total mass of rocket as $m_0=200\text{kg}$and the rate at which fuel is consumed is $\lambda =1\text{kg}/\text{s},$then we have

$m_0-\lambda t=m_0-m_f$

Then we have

$$\begin{gathered} 200-t=200-50 \\ 200-t=150 \end{gathered}$$

Then we obtain

$\begin{array}{rcl}{t}_b& =& 200-150\\ & =& 50\text{sec}\end{array}$

50 seconds required is the time at which all fuel in the rocket is consumed.

Find the velocity

b) Since we know from problem (39), (equation (4)) that the velocity of the rocket at time t is

$v\left(t\right)=0.024t^2+0.2t$

Then we can obtain the velocity at burnout by substituting withseconds into equation (1) as

$\begin{array}{rcl}{v}_b& =& 0.024\times (50{)}^2+0.2\times 50\\ & =& 70\text{m}/\text{s}\end{array}$

Find the height

c) Since we know from problem (39) (equation (6) in problem (39) ) that the height of the rocket at time t is

$s\left(t\right)=0.08t^3+0.1t^2$

Then we can obtain the height at burnout by substituting with$t=50$ seconds into equation (2) as

$\begin{array}{rcl}{s}_b& =& 0.08\times (50{)}^3+0.2\times (50{)}^2\\ & =& 10,500\text{m required}\end{array}$

d) I expected the rocket to attain an altitude higher than burnout height due to thrust.

Differentiate the equation

e) After burnout, differential equation for the rocket in problem (39) becomes

$\frac{dv}{dt}+\frac{k-\lambda }{m_0-50}v=-g+\frac{R}{m_0-50}$

which is

$\begin{array}{rcl}\frac{dv}{dt}+\frac{3-1}{200-50}v& =& -9.8+\frac{2000}{200-50}\\ \frac{dv}{dt}+\frac{2}{150}v& =& -9.8+\frac{2000}{150}\\ \frac{dv}{dt}+\frac{1}{75}v& =& \frac{53}{15}\\ \frac{dv}{dt}& =& \frac{53}{15}-\frac{v}{75}\\ \frac{dv}{dt}& =& \frac{265-v}{75}\end{array}$

Separate the variables

This differential equation is a first order linear and separable D.E, then we can solve it as

$\begin{array}{rcl}\frac{dv}{265-v}& =& \frac{dt}{75}\\ \int \frac{1}{265-v}dv& =& \frac{1}{75}\int dt\\ \int -\frac{1}{v-265}dv& =& \frac{1}{75}\int dt\\ -\ln (v-265)& =& \frac{1}{75}t+c_1\ln (v-265)=-\frac{1}{75}t+c_2\\ e^{\ln \left(v265\right)}& =& e^{\frac{1}{75}t{c}_2}\\ v-265& =& e^{c_2}{e}^{\frac{1}{75}t}\end{array}$

Then we have

$v=c{e}^{\frac{1}{75}t}+265$

Now, to find the value of constantwe have to apply the point of condition$(v,t)=(70\text{m}/\text{s},0),$ then we have

$70=c{e}^0+265$

Then we have

$\begin{array}{rcl}c& =& 70-265\\ & =& -195\end{array}$

After that, substitute with the value ofin equation (3), then we obtain

$v\left(t\right)=-195e^{\frac{1}{75}t}+265$

is the velocity of the rocket after burnout at time t.