Finding the distance traveled by a steamer

[green-beans]

Homework Statement

A steamer starts from rest, the engine exerting a constant propelling force Mf, where M is the mass of the steamer. The resistance of the steamer is assumed to vary as the square of the speed. Show that the distance x traveled in time t is:
x = (V^2 /f) ln cosh (f*t/V ) where V is the full speed of the steamer.

Homework Equations

The Attempt at a Solution

I am not sure whether I understood what is happening in this system correctly but I got the following equation:
Mf - v²M = Ma
f - v² = a
Then, I tried solving it to find x(v) and I obtained:
x = 1/2 ln({f}/{f-V²})
I got a similar expression for v(t):
t = arcsin (V/√f)
But then I am not sure how one an obtain an expression with all three variables in it (i.e. x, V and t)

Thank you in advance!

 

[PeroK]

Homework Statement

A steamer starts from rest, the engine exerting a constant propelling force Mf, where M is the mass of the steamer. The resistance of the steamer is assumed to vary as the square of the speed. Show that the distance x traveled in time t is:
x = (V^2 /f) ln cosh (f*t/V ) where V is the full speed of the steamer.

Homework Equations

The Attempt at a Solution

I am not sure whether I understood what is happening in this system correctly but I got the following equation:
Mf - v²M = Ma
f - v² = a
Then, I tried solving it to find x(v) and I obtained:
x = 1/2 ln({f}/{f-V²})
I got a similar expression for v(t):
t = arcsin (V/√f)
But then I am not sure how one an obtain an expression with all three variables in it (i.e. x, V and t)

Thank you in advance!

Your first two problems are

1) You assumed that the resistance force was $Mv^2$. This is not stated in the problem. It will be $kv^2$ for some constant $k$.

2) You didn't calculate $V$, which is the terminal velocity. It seems that you may be confusing $V$ with the instantaneous velocity $v(t)$

Your dodgy latex isn't helping. These problems are too complicated to try to work with almost unreadable expressions. Try copying some proper latex from other posts.

 

[PeroK]

Let me help you with the latex. You are trying to show: $x(t) = \frac{V^2}{f} \ln (\cosh(\frac{ft}{V}))$

Or, if you want to separate the expression use $ signs:$$x(t) = \frac{V^2}{f} \ln (\cosh(\frac{ft}{V}))$$

If you reply to this, you'll get the latex and you can edit it.

 

[green-beans]

Let me help you with the latex. You are trying to show: $x(t) = \frac{V^2}{f} \ln (\cosh(\frac{ft}{V}))$

Or, if you want to separate the expression use $ signs:$$x(t) = \frac{V^2}{f} \ln (\cosh(\frac{ft}{V}))$$

If you reply to this, you'll get the latex and you can edit it.

I apologise for my poor formatting and thank you - I managed to get the needed expression! However, I was also wondering if I was to find the further distance traveled by a steamer before it comes to rest given its engines are reversed when traveling at full speed, will the equation of motion be as follows:
-kv² - Mf = Ma?
It is just I thought when its engine is reversed, the steamer will still be traveling in positive direction until it comes to rest and the only force that will change its direction is the propelling force produced by the engine.
If the expression is correct, to find the distance I will need to consider x from x=0 to x=X and v from v=V (full speed) to v=0, if I understand this correctly?

 

[PeroK]

I apologise for my poor formatting and thank you - I managed to get the needed expression! However, I was also wondering if I was to find the further distance traveled by a steamer before it comes to rest given its engines are reversed when traveling at full speed, will the equation of motion be as follows:
-kv² - Mf = Ma?
It is just I thought when its engine is reversed, the steamer will still be traveling in positive direction until it comes to rest and the only force that will change its direction is the propelling force produced by the engine.
If the expression is correct, to find the distance I will need to consider x from x=0 to x=X and v from v=V (full speed) to v=0, if I understand this correctly?

Yes, I would assume that the retarding force would be the full engine force plus the velocity related force. And, yes, the steamer will be close enough to $V$ to take that as the initial speed.

Did you manage to solve the first part?

 

[green-beans]

Yes, I would assume that the retarding force would be the full engine force plus the velocity related force. And, yes, the steamer will be close enough to $V$ to take that as the initial speed.

Did you manage to solve the first part?

Yes, I did! After hours of rechecking my work and finding some stupid arithmetic mistakes ! It was painful but I got the result. Thank you so much for your help!