Question

Rocket Motion Suppose a small single-stage rocket of total mass m(t) is launched vertically, the positive direction is upward, the air resistance is linear, and the rocket consumes its fuel at a constant rate. In Problem 22 of Exercises 1.3 you were asked to use Newton’s second law of motion in the form given in (17) of that exercise set to show that a mathematical model for the velocity v(t) of the rocket is given by

$\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{t}}\mathbf{+}\frac{\mathbf{k}\mathbf{-}\mathbf{v}}{{\mathbf{m}}_{\mathbf{0}}\mathbf{-}\mathbf{\lambda }\mathbf{t}}\mathit{v}\mathbf{=}\mathbf{-}\mathit{g}\mathbf{+}\frac{\mathbf{R}}{{\mathbf{m}}_{\mathbf{0}}\mathbf{-}\mathbf{\lambda }\mathbf{t}}$,

where k is the air resistance constant of proportionality, ­ is the constant rate at which fuel is consumed, R is the thrust of the rocket, $\mathit{m}\left(t\right)\mathbf{=}{\mathit{m}}_{\mathbf{0}}\mathbf{-}\mathit{\lambda }\mathit{t}$, ${\mathit{m}}_{\mathbf{0}}$ is the total mass of the rocket at $\mathit{t}\mathbf{=}\mathbf{0}$, and g is the acceleration due to gravity. (a) Find the velocity $\mathit{v}\mathbf{\left(}\mathit{t}\mathbf{\right)}$of the rocket if ${\mathit{m}}_{\mathbf{0}}\mathbf{=}\mathbf{200}\mathbf{}\mathbf{kg}\mathbf{,}\mathit{R}\mathit{=}\mathbf{2000}\mathbf{}\mathit{N}\mathbf{,}\mathit{\lambda }\mathbf{=}\mathbf{1}\mathbf{}\mathbf{kg}\mathbf{/}\mathbf{s}\mathbf{,}\mathbf{}\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}\mathbf{,}\mathit{k}\mathbf{=}\mathbf{3}\mathbf{}\mathit{k}\mathit{g}\mathbf{/}\mathit{s}$and $\mathit{v}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

(b) Use $\mathit{d}\mathit{s}\mathbf{/}\mathit{d}\mathit{t}\mathbf{=}\mathit{v}$and the result in part (a) to ­nd the height s(t) of the rocket at time t .

Short answer

$$\begin{gathered} a\left)v\right(t)=0.024t^2+0.2t \\ b)s\left(t\right)=0.08t^3+0.1t^2 \end{gathered}$$

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) We have a small single-stage rocket is launched vertically upward with air resistance has the first order linear differential equation

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

and we have to obtain its velocity$v\left(t\right)$as the following technique :

Since we have$k=3\text{kg}/\text{s},\lambda =1\text{kg}/\text{s},m_0=200\text{kg},R=2000\text{N}$and$g=9.8\text{m}/{\text{s}}^2$, then we can have equation (1) as

$\begin{array}{rcl}\frac{dv}{dt}+\frac{3-1}{200-\left(1\right)t}v& =& -9.8+\frac{2000}{200-\left(1\right)t}\\ \frac{dv}{dt}+\frac{2}{200-t}v& =& -9.8+\frac{2000}{200-t}\\ \frac{dv}{dt}+\frac{2}{200-t}v-\frac{2000}{200-t}& =& -9.8\\ \frac{dv}{dt}+\frac{2}{200-t}(v-1000)& =& -9.8\end{array}$

This differential equation is a first order differential equation as the form

$\frac{dv}{dt}+p\left(t\right)(v-c)=g\left(t\right)$

Integrating factor

then we can solve it as the following:

We have to obtain an integrating factorfor this D.E as

$\begin{array}{rcl}F& =& e^{\int p\left(t\right)dt}\\ & =& e^{\int \frac{2}{200-t}dt}\\ & =& e^{2\int \frac{-1}{200-t}dt}\\ & =& e^{2\ln \left(200t\right)dt}\\ & =& \frac{1}{{(200-t)}^2}\end{array}$

After that, multiply both sides of first order differential equation shown in (2) by this integrating factor, then we have

$\begin{array}{rcl}\frac{1}{{(200-t)}^2}\frac{dv}{dt}+\left(\frac{1}{{(200-t)}^2}\right)\frac{2}{200-t}(v-1000)& =& -9.8\left(\frac{1}{{(200-t)}^2}\right)\\ \frac{1}{{(200-t)}^2}\frac{dv}{dt}+\frac{2}{{(200-t)}^3}(v-1000)& =& -\frac{9.8}{{(200-t)}^2}\end{array}$

Then if we undo one step, we can have

$\frac{d}{dt}\left(\frac{1}{{(200-t)}^2}(v-1000)\right)=-\frac{9.8}{{(200-t)}^2}$

Then we can do integration as

$\begin{array}{rcl}\int d\left(\frac{1}{{(200-t)}^2}(v-1000)\right)& =& -9.8\int \frac{1}{{(200-t)}^2}dt\\ \frac{1}{{(200-t)}^2}(v-1000)& =& -9.8\times \frac{1}{200-t}+c\\ v-1000& =& -9.8\frac{{(200-t)}^2}{200-t}+c(200-t{)}^2\end{array}$

Then we have

$v\left(t\right)=-9.8(200-t)+c(200-t{)}^2+1000$

Find the value of constant c and k, substitution

Now, to find the value of constant c, we have to apply the point of condition

$(v,t)=(0,0)$, then we have

$$\begin{gathered} 0=-9.8(200-0)+c(200-0{)}^2+1000 \\ 0=-960+40000c \end{gathered}$$

Then we have

$\begin{array}{rcl}c& =& \frac{960}{40000}\\ & =& 0.024\end{array}$

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

$\begin{array}{rcl}v\left(t\right)& =& 0.024(200-t{)}^2-9.8(200-t)+1000\\ & =& 960-9.6t+0.024t^2-1960+9.8t+1000\\ & =& 0.024t^2+0.2t\text{required}\end{array}$

is the velocity of the rocket at time t.

b) After that, since we have the relation betweenandas

$v=\frac{ds}{dt}$

Then by separation for variables and integration, we can have

$\begin{array}{rcl}\int ds& =& \int vdt\\ s& =& \int vdt\end{array}$

Then by substituting withfrom equation (4) into this equation, we can have

$\begin{array}{rcl}s& =& \int \left(0.024t^2+0.2t\right)dt\\ & =& \int 0.024t^{2}dt+\int 0.2tdt\\ & =& \frac{0.024}{3}{t}^3+\frac{0.2}{2}{t}^2+k\\ & =& 0.08t^3+0.1t^2+k\end{array}$

Now, to find the value of constant k, we have to apply the point of condition$\left(s,t\right)=\left(0,0\right),$ then we have

$0=0.08(0{)}^3+0.1(0{)}^2+k$

Then we have

$k=0$

After that, substitute with the value ofin equation (5), then we obtain

$s\left(t\right)=0.08t^3+0.1t^2\text{required}$

is the height of the rocket at time t.

$$\begin{gathered} a\left)v\right(t)=0.024t^2+0.2t \\ b)s\left(t\right)=0.08t^3+0.1t^2 \end{gathered}$$