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dynamics

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 travelled in time t is

x=[(V^2)/f]ln cosh [(ft)/V]

where V is the full speed of the steamer. Hence show for large values of t

x=Vt - [(V^2)/f]ln2.

If, when the steamer is travelling at full speed, the engines are reversed, show that the steamer will come to rest after travelling a further distance [(V^2)/2f]ln2.

How can this be done?
Reply 1
Avatar for teachercol
teachercol
9
Write down an equation for the force on the steamer.

Then get an expression for the acceleration.

Integrate it twice to get displacment.
Reply 2
Avatar for gavak
gavak
OP
I've tried with:
F=Mf
Resistence= k |v|^2 where k is a costant of proportionality
So a = dv/dt = Mf - k|v|^2
But then I stuck with the integration.
I think this is not correct..
Can you explain it to me?
Cheers.
Reply 3
Avatar for teachercol
teachercol
9
F = Mf - kv^2

Divide by m

a = dv / dt = f - (k/m) v^2 = f - k' v^2

Now separete variables and integrate wrt to t - if you cant do this integral, then the problem is maths not physics!
Reply 4
Avatar for gavak
gavak
OP
What does it mean that V is the full speed of the steamer?
Reply 5
Avatar for teachercol
teachercol
9
Maximum speed ( when kv^2 = Mf)
Reply 6
Avatar for gavak
gavak
OP
:frown: Any additional hint?

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