# Boat Drag Force

I have a question about a physics problem, this comes from Barger's Classical Mechanics: A boat is slowed by a drag force, F(v) and its velocity decreases according to the formula $$v=c^2(t-t_{1})^2$$ where c is a constant and $$t_{1}$$ is the time at which it stops. Find the force F(v) as a function of v.

So I figured $$F=ma=mdv/dt=md/dt(c^2(t-t_1)^2)$$. After differentiation, I get $$F=2mc\sqrt{v}$$. The solution in the back has a negative sign, and I'm wondering what justifies the negative. I'm assuming it has somethign to do with the fact that it is a drag force and the velocity is decreasing, but it's not entirely clear. Thanks in advance.

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Because in your original problem the velocity is decreasing, thus the first derivative of velocity
must be negative right?

SGT
For the force to slow down the boat, it must be applied in the inverse sense of the velocity. If the velocity is positive, the force must be negative.

It's obvious in the given equation...

If you were to look at the given equation the (t-t1)^2 part, you would notice that as time increases from 0, the velocity becomes less negative until eventually reaching 0 when t=t1. This must mean that velocity is directed in the negative direction and force is in the positive direction.

If however you were to take it as (t1 - t)^2 , which wouldn't really change the velocity equation, the directions are reversed and force is in the negative direction.

Hope this helped.

Doc Al
Mentor
rahullak said:
If you were to look at the given equation the (t-t1)^2 part, you would notice that as time increases from 0, the velocity becomes less negative until eventually reaching 0 when t=t1. This must mean that velocity is directed in the negative direction and force is in the positive direction.
The velocity is always positive, the acceleration negative. As SGT (and jaredkipe) stated, the force must be negative.

When taking the square root of v, there are two answers: one plus, one minus. You have to choose the minus answer, since that's the one that fits this situation.

Just to be clear:
$$c (t - t_1) = -\sqrt{v}$$
$$c (t - t_1) \ne +\sqrt{v}$$

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