# Physics Problem: Canoe Velocity & Distance | Viet Dao

• VietDao29
In summary, a canoe with a mass of 0.5kg has an initial velocity of 10 m/s and is subjected to a decelerating force of 0.5v as it moves along a river. The distance it will travel when its velocity is 5m/s is 5m, and when it stops it will have traveled 10m. To solve for this, the equation 0.5\frac{dv}{dt}=-0.5 v(t) must be integrated, using the initial condition given in the problem. The correct solution is v(t)=\bar{C}e^{-t}, where \bar{C} is found by imposing the initial condition.
VietDao29
Homework Helper
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,

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

Daniel.

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

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

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

Daniel.

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,

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.

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

## 1. How do you calculate the velocity of a canoe?

The velocity of a canoe can be calculated by dividing the distance traveled by the time it took to travel that distance. The formula for velocity is velocity = distance/time.

## 2. What factors affect the velocity of a canoe?

The velocity of a canoe can be affected by various factors such as the force of the paddling, the weight of the canoe and passengers, the wind and water currents, and the shape and design of the canoe.

## 3. What is the relationship between velocity and distance for a canoe?

The relationship between velocity and distance for a canoe is directly proportional. This means that as the velocity increases, the distance traveled also increases, and vice versa.

## 4. How does the mass of the canoe and passengers affect the velocity?

The mass of the canoe and passengers can affect the velocity by adding more weight and increasing the resistance against water. This can decrease the velocity of the canoe, requiring more force to maintain a certain speed.

## 5. How can the velocity of a canoe be increased?

The velocity of a canoe can be increased by reducing the weight of the canoe and passengers, paddling with more force and efficiency, and taking advantage of favorable wind and water currents. Improving the design and shape of the canoe can also help increase its velocity.

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