Motion with a changing acceleration

AI Thread Summary
The discussion focuses on calculating the time it takes for a boat to decelerate from an initial speed v_i to a final speed v_f after its engine is turned off, with the friction force f being proportional to its speed. Participants suggest using calculus to derive the relationship between force, mass, and acceleration, noting that the acceleration is not constant due to the varying friction force. The key equation derived is a = (kv)/m, leading to the integration of time with respect to velocity. The final expression for time t is proposed as t = (m/k) * ln(v_f/v_i). This approach effectively combines concepts of motion and calculus to solve the problem.
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Homework Statement


A boat, whose mass is m, is traveling at a speed of v_i when its engine is shut off. The magnitude of the friction force f between boat and water is proportional to the speed v of the boat: f=kv. Find the time t for the boat to slow down to speed v_f

Homework Equations


The Attempt at a Solution


I think there should be some transformations with the help of calculus, but I really don't have any ideas.
 
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fizics said:

Homework Statement


A boat, whose mass is m, is traveling at a speed of v_i when its engine is shut off. The magnitude of the friction force f between boat and water is proportional to the speed v of the boat: f=kv. Find the time t for the boat to slow down to speed v_f


Homework Equations





The Attempt at a Solution


I think there should be some transformations with the help of calculus, but I really don't have any ideas.

Start by writing the general equation of motion relating F, m and a. Now, you have an additional equation above due to the varying resistance force as a function of v. These two equations are related.

Now look at the general kinematic equations of motion in a constant acceleration with the usual forms of v= and x= ... Those are simplifications where integrations had simple answers because the acceleration was constant. Re-write the general forms of those equations, and look at what changes when the acceleration is not constant.

That should get you started. Show us your calculations as you work through this...
 
f= ma= kv, so a= (kv)/m. Thus,
v_f - v_i = ∫a dt (from 0 to t) = (k/m)* ∫v dt (from 0 to t) = ?
 
Hi,

F=kv=ma

\frac{dv}{dt}=\frac{Kv}{m}

dt=\frac{m}{kv} dv

t=\int_{vi}^{vf} \frac{m}{kv} \ \ dv

since K and m are constants therefore

t=\frac{m}{k} \int_{vi}^{vf} \frac{1}{v} \ \ dv

then you have to integrate it...

right?
 
Bright Wang said:
Hi,

F=kv=ma

\frac{dv}{dt}=\frac{Kv}{m}

dt=\frac{m}{kv} dv

t=\int_{vi}^{vf} \frac{m}{kv} \ \ dv

since K and m are constants therefore

t=\frac{m}{k} \int_{vi}^{vf} \frac{1}{v} \ \ dv

then you have to integrate it...

right?

I don't have the answer but I think it's right. How do you come up with this? Is it because in F=kv=ma, there's v, and you want to have a dv in the expression?
 
hmm you said that... "f= ma= kv"

basicly you use the fact
a=\frac{dv}{dt}

can you integrate it and get the equation?
 
t= (m/k)*ln(v_f/v_i)
 
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