1. ## rocket.

The engines of a rocket of mass 20 tonnes produce a thrust of 398.5 KN The rocket is Launched in a vertical direction from the surface of the earth. The air resistance is v^2 N where v is the speed of the rocket. find its speed after a minute and also find the max speed which the rocket tends to.

This is what I tried.

so using F=ma

I let 398500-v^2 =20000a

a=(398500-v^2)/20000

so dv/dt = (398500-v^2)/20000

I found this difficult to integrate but rearranged to get

dt(1/20000) = dv(1/398500)

I then integrated this to get

t20000^-1 = [ln(398500-v^2)]/-2v

this is unsolvable.

2. ## Re: rocket.

Few things. A mass of 20 tonnes is affected by gravity as well as air resistance. Are you assuming that gravity is negligible? Note: The higher the rocket gets, the lower the gravitational constant becomes, and the more complicated the differential equation becomes, so I completely understand if you are ignoring gravity, but it does need to be stated.

Now, you go from:

$\dfrac{dv}{dt} = \dfrac{398500-v^2}{20000}$

to:

$\dfrac{dt}{20000} = \dfrac{dv}{398500}$

You seem to have lost the $-v^2$.

The correct differential equation is:

$\dfrac{dt}{20000} = \dfrac{dv}{398500-v^2}$

Integrating both sides, you get:

$\displaystyle \int \dfrac{dt}{20000} = \int \dfrac{dv}{398500-v^2}$

You got the LHS correct, but I have no idea how you got the RHS.

Let $v = \sqrt{398500} \sin \theta, dv = \sqrt{398500} \cos \theta d\theta$

Then, the integral becomes:

\begin{align*}\dfrac{t}{20000} + C & = \int \dfrac{\sqrt{398500} \cos \theta d \theta }{398500(1-\sin^2 \theta)} \\ & = \int \dfrac{ \sec \theta d\theta }{\sqrt{398500}} \\ & = \dfrac{1}{\sqrt{398500}} \ln \left| \sec \theta + \tan \theta \right|\end{align*}

You have $\sin \theta = \dfrac{v}{\sqrt{398500}} = \dfrac{\text{opp}}{\text{hyp}}$

This means $\tan \theta = \dfrac{\text{opp}}{\text{adj}} = \dfrac{v}{\sqrt{398500-v^2}}, \sec \theta = \dfrac{\text{hyp}}{\text{adj}} = \dfrac{\sqrt{398500}}{\sqrt{398500-v^2}}$

Plugging in, you get:

$\dfrac{t}{20000} + C = \dfrac{1}{\sqrt{398500}}\ln \left|\dfrac{v}{\sqrt{398500-v^2}}+\dfrac{\sqrt{398500}}{\sqrt{398500-v^2}}\right|$

At $t=0$, you have $v=0$, so, let's solve for $C$:

$C = \dfrac{1}{\sqrt{398500}}\ln \left| 0 + 1 \right| = 0$

So, your final formula relating the two is given above. Next, we plug in for $t=1 \text{ min}$. Since Newtons are $\dfrac{\text{kg}\cdot \text{m}}{\text{s}^2}$, our time is in seconds, so one minute is 60 seconds. Plug in 60 for $t$.

$e^{\tfrac{60\sqrt{398500}}{20000}} = \dfrac{v + \sqrt{398500}}{\sqrt{398500-v^2}}$

Square both sides:

$e^{\tfrac{60\sqrt{398500}}{10000}} = \dfrac{v^2+2v\sqrt{398500} + 398500}{398500-v^2}$

To simplify the calculations, let $k = e^{\tfrac{60\sqrt{398500}}{10000}}$

$0 = (1+k)v^2+2v\sqrt{398500} + 398500(1-k)$

Now, you have a quadratic in terms of $v$. Solve that and take the positive value for $v$ (I assume one value will be positive and one will be negative).

Next, to solve for the maximum velocity, you want $\dfrac{dv}{dt} = 0$. So, we return to your original equation and plug that in. This gives a critical point for velocity at $v=\sqrt{398500}$. We can see that when $v$ is less that that, the derivative is positive, when it is greater, it is negative. So, this critical value occurs at a maximum, and $v=\sqrt{398500}$ is, in fact, the maximum.

3. ## Re: rocket. Originally Posted by edwardkiely The engines of a rocket of mass 20 tonnes produce a thrust of 398.5 KN The rocket is Launched in a vertical direction from the surface of the earth. The air resistance is v^2 N where v is the speed of the rocket. find its speed after a minute and also find the max speed which the rocket tends to.

This is what I tried.

so using F=ma

I let 398500-v^2 =20000a

a=(398500-v^2)/20000

so dv/dt = (398500-v^2)/20000
Yes, that is correct so far.

I found this difficult to integrate but rearranged to get

dt(1/20000) = dv(1/398500)
What happened to the v^2?

I then integrated this to get

t20000^-1 = [ln(398500-v^2)]/-2v

this is unsolvable.
It looks like you integrated wrong!

Starting from $dv/dt= (398500-v^2)/20000$
Yes, you can write it as $\frac{dv}{398500- v^2}= 20000dt$

$398500- v^2= (\sqrt{398500}- v)(\sqrt{398500}+ v)$.

Using "partial fractions" $\frac{1}{\sqrt{398500}}- v^2$ can be written as $\frac{1}{2\sqrt{398500}}\left(\frac{1}{\sqrt{3985 00}+ v}+ \frac{1}{\sqrt{398500}- v}\right)$.

Then integrating the left side gives $\frac{1}{2\sqrt{398500}}\left(ln(\sqrt{398500}+ v)- ln(\sqrt{398500}- v)\right)$$= frac{1}{2\sqrt{398500}}ln\left(\frac{\sqrt{398500} + v}{\sqrt{398500}- v}\right)$

So we have $frac{1}{2\sqrt{398500}}ln\left(\frac{\sqrt{398500 }+ v}{\sqrt{398500}- v}\right)= 20000t+ C$ where "C" is the combined constants of integration. We know that when t= 0, v= 0 so $C= \frac{1}{2\sqrt{398500}}ln(1)= 0$. We have $frac{1}{2\sqrt{398500}}ln\left(\frac{\sqrt{398500 }+ v}{\sqrt{398500}- v}\right)= 20000t$.

That certainly can be solved for v! To solve that for v, first multiply both sides by $2\sqrt{398500}$ to get $ln\left(\frac{\sqrt{398500}+ v}{\sqrt{398500}- v}\right)= 40000\sqrt{398500}t$.

Get rid of the logarithm by taking the exponential of both sides:
$\frac{\sqrt{398500}+ v}{\sqrt{398500}- v}= e^{40000\sqrt{398500}t}$

Multiply on both sides by $\sqrt{398500- v}$:
$\sqrt{398500}+ v= (\sqrt{398500}- v)e^{40000\sqrt{398500}t}= \sqrt{398500}e^{40000\sqrt{398500}t}- ve^{40000\sqrt{398500}t}$.

Add $ve^{40000\sqrt{398500}t}$ to both sides and subtract $\sqrt{398500}$:
$\left(1+ e^{40000\sqrt{398500}t}\right)v= \sqrt{398500}\left(e^{40000\sqrt{398500}t}- 1\right)$.

Finally, divide both by $1+ e^{40000\sqrt{398500}t}$:

$v= \sqrt{398500}\left(\frac{e^{40000\sqrt{398500}t}- 1}{e^{40000\sqrt{398500}t}+ 1}\right)$

4. ## Re: rocket.

I do not think yous are getting the right answer. The answer at the back of the book is 393.3 m/s.

I think I was supposed to take gravity into account. we re supposed to assume gravity is constant.so using F=ma it should of read 398500-v^2 -20000g=20000a

so a=(398500-v^2 -20000g)20000

rearrange this and yous will get [1/(202500-v^2) ]dv =1/20000^-1t

http://calculusworld.weebly.com/uplo...gif?1348107889 using the third one from the top ...

integrate this and put in the limits and solve for v and I get v=393.3

so it was quite easy.

I am not sure how to do the last part though. what is the max speed the rocket tends to? the answer at the back of the book is 450.

5. ## Re: rocket. Originally Posted by edwardkiely I do not think yous are getting the right answer. The answer at the back of the book is 393.3 m/s.

I think I was supposed to take gravity into account. we re supposed to assume gravity is constant.so using F=ma it should of read 398500-v^2 -20000g=20000a

so a=(398500-v^2 -20000g)20000

rearrange this and yous will get [1/(202500-v^2) ]dv =1/20000^-1t

http://calculusworld.weebly.com/uplo...gif?1348107889 using the third one from the top ...

integrate this and put in the limits and solve for v and I get v=393.3

so it was quite easy.

I am not sure how to do the last part though. what is the max speed the rocket tends to? the answer at the back of the book is 450.
Read my solution from Part #2. I specifically ask about gravity. I also tell you exactly what to do to find the maximum velocity. Just because the specifics change does not mean that the method does. Follow the method I gave you, and it will be easy to see how the maximum velocity is 450.

$a = \dfrac{398500-v^2-20000g}{20000}$

At $a=0$, you have $v = \sqrt{398500-20000g} = 450$

6. ## Re: rocket.

sorry, forgot to read the end of your post