Chapter 8, Problem 8.1P

To determine

The time for which rocket fires so that the sled covers 2000 m.

Answer to Problem 8.1P

The time taken by the rocket sled to cover 2000 meters is $t=18.028\sec$.

Explanation of Solution

Given:

The equation of motion for the rocket sled is,

$6\dot{v}=2700-24v$

$v\left(0\right)=0$

Concept Used:

Use Laplace transform to obtain the response $v\left(t\right)$ from the given equation of motion.

Calculation:

Since,

$6\dot{v}=2700-24v$

On taking Laplace of the above equation.

$\begin{array}{l}6\dot{v}+24v-2700=0\\ \Rightarrow 6\left(sV\left(s\right)-v\left(0\right)\right)+24V\left(s\right)=\frac{2700}{s}\\ \Rightarrow V\left(s\right)=\frac{2700}{s\left(6s+24\right)}=\frac{450}{s\left(s+4\right)}\end{array}$

On using partial fraction expansion,

$V\left(s\right)=\frac{450}{s\left(s+4\right)}=112.5\left(\frac{1}{s}-\frac{1}{s+4}\right)$

Taking inverse Laplace transform of this,

$v\left(t\right)=112.5\left(1-e^{-4t}\right)$

Since,

$v\left(t\right)=\frac{dx\left(t\right)}{dt}$

Therefore,

$\begin{array}{l}\frac{dx\left(t\right)}{dt}=112.5\left(1-e^{-4t}\right)\\ \Rightarrow dx\left(t\right)=112.5\left(1-e^{-4t}\right)dt\end{array}$

On taking integration both sides,

$\begin{array}{l}\Rightarrow {\displaystyle \int dx\left(t\right)}={\displaystyle \int 112.5\left(1-e^{-4t}\right)dt}\\ \Rightarrow x\left(t\right)=112.5\left(t+\frac{1}{4}{e}^{-4t}\right)+C\end{array}$

At $t=0$,

$\begin{array}{l}x\left(0\right)=112.5\left(0+\frac{1}{4}{e}^0\right)+C\\ \Rightarrow \frac{112.5}{4}+C=0\\ \Rightarrow C=-28.125\end{array}$

Now, we have the displacement response for the system as follows:

$x\left(t\right)=28.125e^{-4t}+112.5t-28.125$

Here, the time constant for the response is $\tau =\frac{1}{4}\sec$.

Thus, the response reaches its steady state at $t=4\tau$ since $e^{-4t}<0.02$ at $t=4\tau$.

At $t=4\tau$,

$\begin{array}{l}x\left(t\right)=28.125e^{-4t}+112.5t-28.125\\ \Rightarrow x\left(4\tau \right)=28.125e^{-4}+112.5\times 1-28.125\\ \Rightarrow x\left(4\tau \right)=84.890\text{m}\end{array}$

This is very less than the distance 2000 meters. Therefore, for calculating the time taken by the sled to cover 2000 meter the exponential part could be neglected.

Thus,

$\begin{array}{l}x\left(t\right)=112.5t-28.125\\ \Rightarrow 2000=112.5\times t-28.125\\ \Rightarrow t=\frac{2028.125}{112.5}=18.028\sec \end{array}$.

Conclusion:

The time taken by the rocket sled to cover 2000 meters is $t=18.028\sec$.