Physics Problem: Canoe Velocity & Distance | Viet Dao

[VietDao29]

Hi,
I just want to ask you guys a problem I had at the very beginning of grade 10th. I am going to go to grade 11th next year. But I still cannot figure this out. (So it's obviously not homework). Any help will be highly appreciated.
A canoe (mass 0.5kg) has an initial velocity of 10 m/s. As it moves along the river, a force F = 0.5v acts on it (to make it decelerate). Calculate the distance the canoe will go when:
a/ Its velocity is 5 m/s.
b/ The canoe stops.
(I'm sorry if my translation of the problem is not clear).
The book gives the answer for a is 5m, and b is 10m.
-----
So what I have so far is a = -v. And I know that to calculate the distance, I must have a function v with respect to t. But... how can I have it?
Thanks a lot.
Viet Dao,

 

[dextercioby]

Can u integrate that ODE...?It's separable.

Daniel.

 

[VietDao29]

Can you give me a little hint (just a little bit to clear my mind)? I am stuck...
Viet Dao,

 

[dextercioby]

0.5\frac{dv}{dt}=-0.5 v(t)

is the separable I-st order ODE that needs to be solved,using the initial condition the problem's giving.

Daniel.

 

[VietDao29]

Okay, I'm not sure. But here's what I get:
\frac{dv}{dt} = -v(t)
And I want
s = \int^\varepsilon_\xi v(t)dt
Plug that in the function and I have:
s = \int^\alpha_\beta -1 dv
s = \int^\beta_\alpha 1 dv
For a/
s = \int^{10}_{5} 1 dv = 5
For b/
s = \int^{10}_{0} 1 dv = 10
Is it correct? Is there another way?
Thanks,
Viet Dao,

 

[dextercioby]

It's not correct.There's another way.It's called separation of variables.

\frac{dv}{v}=-dt (1)

\int \frac{dv}{v}=-\int dt (2)

\ln v(t)=-t+C (3)

v(t)=\bar{C}e^{-t} (4)

,where \bar{C}=e^{C} (5)

(4) is the solution to your problem.You need to find the integration constant by imposing the initial condition.

Daniel.

 

[VietDao29]

Thanks very much. Finally, I know how to solve this problem.
Viet Dao,