# Thunderbolt (Viva Rocketry! Part 1.1)

Fig. 1

Shown in Fig. 1 is an experimental car propelled by a rocket motor. The drag force (air resistance) is given by $R = \beta v^2$. The initial mass of the car, which includes fuel of mass $m_f$, is $m_0$. The rocket motor is burning fuel at the rate of $q$ with an exhaust velocity of $u$ relative to the car.  The car is at rest at $t=0$. Show that the velocity of the car is given by, for $0 \leq t \leq T$,

$v(t) = \mu \cdot \frac{1-({m \over m_0})^{\frac{2\beta \mu}{q}}}{1+({m \over m_0})^{\frac{2\beta \mu}{q}}}$,

where $m=m_0-qt, \mu^2=\frac{qu}{\beta}$, and $T=\frac{m_f}{q}$ is the time when the fuel is burnt out.

We have derived the governing equation of rocket flight in “Viva Rocketry! Part 1“, namely,

$F = m \frac{dv}{dt} + u \frac{dm}{dt}\quad\quad\quad(1)$

From $m=m_0-qt$,  we have

$\frac{dm}{dt} = -q$.

Apply air resistance $R=\beta v^2$ as the external force, (1) becomes

$-\beta v^2 = (m_0 - q t) \frac{dv}{dt} - u q$.

And the car is at rest initially implies

$v(0)=0$.

It follows that the motion of the car can be modeled by an initial-value problem

$\begin{cases} -\beta v^2 = (m_0 - q t) \frac{dv}{dt} - u q \\ v(0) = 0 \end{cases}\quad\quad\quad(2)$

It suffices to show that the given $v(t)$ is the solution to this initial-value problem:

Fig. 2

An alternative is obtaining the stated $v(t)$ through solving (2).

Fig. 3

The fact that $m = m_0 -qt, (-1)^{2 \frac{\sqrt{b}\sqrt{u}}{\sqrt{q}}} = 1$ simplifies the result considerably,

$\frac{\sqrt q \sqrt u}{\sqrt \beta}\cdot \frac{m_0^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}-m^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}{m_0^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}-m^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}\quad\quad\quad(3)$

Divide both the numerator and denominator of (3) by $m_0^{\frac {2 \sqrt{\beta} \sqrt{u}}{\sqrt{q}}}$ then yields

$\frac{\sqrt q \sqrt u}{\sqrt \beta}\cdot \frac{1-(\frac{m}{m_0})^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}{1-(\frac{m}{m_0})^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}$

which is equivalent to the given $v(t)$ since $\mu^2=\frac{qu}{\beta}$.

At time $t=T$, the fuel is burnt out. It means

$m_0-m_f = m_0 - qT$.

Therefore,

$T = \frac{m_f}{q}$

Exercise 1: Solve (2) manually.

Hint: The differential equation of (2) can be written as $\frac{1}{uq - \beta v^2} \frac{dv}{dt} = \frac{1}{m_0 - q t}$.

Exercise 2: For $m_0=900\;kg, m_f=450\;kg, q=15\;kg/sec, u=500\;m/sec, \beta=0.3$, what is the burnout velocity of the car?