When will the fishing boat and the speedboat meet?

In summary: You know that the average value of the acceleration is going to be positive, right? So, using the average value of the acceleration would imply that the velocity will keep increasing with time, while we know that it won't.Now you could use the average value of the acceleration to find the average velocity, and that would give you a good idea of how fast the boat is moving. But since we know the acceleration is a function of time, we can't use the average value.Also, I'm not sure how you would go about using the second fundamental theorem of calculus in this problem. Can you elaborate a bit more?In summary, the problem involves a fishing boat and a speedboat passing the same point at the same time,
  • #1
kitsh
7
0

Homework Statement


a)
A fishing boat of mass m moves through water in the +x direction. At time t = 0, it is at location x = 0 and has speed v0 at that precise moment, the boat’s captain turns off the engines and let's the boat drift. Taking the drag force into account, Fd=-Kv^2, calculate the boats velocity as a function of time.

b)
A sleek speedboat also passes the point x = 0 at time t = 0. Its speed at that moment is only half the fishing boat’s speed: v0/2. Now this super-sleek speedboat is designed so well that it experiences a negligible amount of drag: Fd=0 for the speed boat. Consequently, the speedboat’s captain doesn’t even
have his engines turned on! The fishing boat passes the speedboat, but eventually, the speedboat will catch up to the fishing boat. At what position X do the two boats meet? Write down a formula for X that is solvable and involves only known parameters. (Don’t try to solve it, it doesn’t have an analytic solution.)

Homework Equations


F=ma

The Attempt at a Solution


For part a I set the drag force equal to ma and solved it to be v(t)=v0-m/kt. I don't know if this is right but it makes sense to me, I have no idea how to approach part b though.
 
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  • #2
kitsh said:
For part a I set the drag force equal to ma and solved it to be v(t)=v0-m/kt.

You were right to solve for acceleration in ma = -Kv2 and plug your result into the equation v = v0 + at. However, you might want to do the algebra again, as I do not think that your function is correct.
 
  • #3
Also, make sure to use parentheses to denote when you have a fraction multiplied by another value. For example, 1/2x is very different from (1/2)x.
 
  • #4
kitsh said:
For part a I set the drag force equal to ma and solved it to be v(t)=v0-m/kt

Why don't you try solving this problem using relative motion? Moreover I think the problem will get a lot simpler if you could relate boat's velocity with displacement. And I think you need to recheck your expression.
 
  • #5
kitsh said:
Now this super-sleek speedboat is designed so well that it experiences a negligible amount of drag: Fd=0 for the speed boat. Consequently, the speedboat’s captain doesn’t even
have his engines turned on!
I think according to the question the acceleration of second boat (the sleek speedboat) is zero.
 
  • #6
Hello kitsh,

Welcome fo Physics Forums! :smile:

kitsh said:

Homework Statement


a)
A fishing boat of mass m moves through water in the +x direction. At time t = 0, it is at location x = 0 and has speed v0 at that precise moment, the boat’s captain turns off the engines and let's the boat drift. Taking the drag force into account, Fd=-Kv^2, calculate the boats velocity as a function of time.

b)
A sleek speedboat also passes the point x = 0 at time t = 0. Its speed at that moment is only half the fishing boat’s speed: v0/2. Now this super-sleek speedboat is designed so well that it experiences a negligible amount of drag: Fd=0 for the speed boat. Consequently, the speedboat’s captain doesn’t even
have his engines turned on! The fishing boat passes the speedboat, but eventually, the speedboat will catch up to the fishing boat. At what position X do the two boats meet? Write down a formula for X that is solvable and involves only known parameters. (Don’t try to solve it, it doesn’t have an analytic solution.)

Homework Equations


F=ma

The Attempt at a Solution


For part a I set the drag force equal to ma and solved it to be v(t)=v0-m/kt. I don't know if this is right but it makes sense to me, I have no idea how to approach part b though.
I got a different answer.

The way you typed in your answer is kind of ambiguous but any way you interpret it, something is not quite right. If you meant,
[tex] v(t) = v_0 -\frac{m}{k}t [/tex]
this has problems that it implies that the acceleration is uniform, which it is not. The acceleration is not uniform in this problem.

If you meant
[tex] v(t) = v_0 -\frac{m}{kt} [/tex]
This is not right either, since the velocity blows up for small t.

The crucial check for your answer is to put your expression for velocity back into the differential equation and see if checks out.

I'd elaborate more, but you haven't indicated your differential equation yet. And by the way, there is a differential equation involved with this problem that must be solved.
 
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  • #7
AlephNumbers said:
You were right to solve for acceleration in ma = -Kv2 and plug your result into the equation v = v0 + at. However, you might want to do the algebra again, as I do not think that your function is correct.

That's not right either (sorry). Yes, it's true that [itex] ma = -kv^2 [/itex], but it's not true that [itex] v = v_0 +at [/itex]. The latter equation is fine for uniform acceleration, but does not apply here. In this problem the fishing boat's velocity and acceleration are both functions of time.
 
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  • #8
Hint: You can solve this problem by noting that [itex] a = \frac{dv}{dt} [/itex]. Express your [itex] ma = -kv^2 [/itex], in that form. You should find that using the "separation of variables" method of solving differential equations comes in handy here. Calculus is necessary.
 
  • #9
Actually, there is a way to solve this without differential equations or advanced calculus. You can use the second fundamental theorem of calculus to find the average value of the acceleration function over the interval v - v0, which can then be substituted into the equation v = vo + at.

Actually, that that really doesn't help with the first part of the question very much, does it?
 
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  • #10
AlephNumbers said:
Actually, there is a way to solve this without differential equations or advanced calculus. You can use the second fundamental theorem of calculus to find the average value of the acceleration function over the interval v - v0, which can then be substituted into the equation v = vo + at.
I question whether that would work.

You see, the equation (that we've all come to love in first year, undergrad [or perhaps high school] physics) [itex] v = v_0 + at [/itex] is itself a solution to a differential equation. It's not magic; rather it is the solution to a differential equation itself, but a simpler one. My point is that it's a different, differential equation than the one that applies here. The [itex] v = v_0 + at [/itex] equation assumes by its very nature that acceleration is uniform.

Allow me to explain. Suppose from the start we assume that [itex] a [/itex] is constant. We start with, [itex] \frac{dv}{dt} = a [/itex]. Then we can write, [itex] dv = adt [/itex]. When integrating, since we know [itex] a [/itex] is constant, we can pull it out of the integral, [itex] \int dv = a \int dt [/itex]. After integrating we have [itex] v = at + C [/itex], applying our initial conditions we find knowing that [itex] v(0) = v_0 [/itex], meaning that [itex] C = v_0 [/itex] we end up with the equation for uniform acceleration, [itex] v = v_0 + at [/itex]

But that whole equation relies on being able to pull the constant [itex] a [/itex] out from under the integral. It means the equation simply does not apply when acceleration is not uniform, and cannot be used when acceleration varies with time.
 
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  • #11
For the problem at hand in this thread, we can qualitatively infer the requirements of an appropriate solution.
  • [itex] v [/itex] needs to equal [itex] v_0 [/itex] when [itex] t = 0[/itex].
  • As [itex] t [/itex] approaches infinity, the velocity should trail off to zero. The velocity should approach zero with increasing time, but should never quite reach zero. If the velocity turns negative (thus turning the boat around) something is wrong.
  • And of course ultimately, if you take the derivative of the velocity function, [itex] \frac{d}{dt} \left\{ v(t) \right\} [/itex] and we take the square of the velocity function, [itex] \left( v(t) \right)^2 [/itex], they must satisfy [itex] ma = -kv^2 [/itex]. With more detail that's rewritten as [itex] m \frac{d}{dt} \left\{ v(t) \right\} = -k \left( v(t) \right)^2 [/itex].
So once a solution is solved for, one can use these sanity checks to make sure it makes sense.
 
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  • #12
AlephNumbers said:
Actually, there is a way to solve this without differential equations or advanced calculus. You can use the second fundamental theorem of calculus to find the average value of the acceleration function over the interval v - v0, which can then be substituted into the equation v = vo + at.

I cannot recall any method to solve this differential equation without integration. Cab you please site the exact method and if possible an example.
 
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  • #13
The method was mentioned above - the method of separation of variables; a common technique for solving certain types of differential equations.
 
  • #14
Vatsal Sanjay said:
I cannot recall any of "advance calculus" methods to solve this differential equation. Can you please site the exact method and if possible an example.
It's not particularly advanced. As collinsmark posted, your differential equation is ##m\frac{dv}{dt}=-kv^2##. Rearrange that to get all the references to v on one side and the reference to t on the other, then integrate.
 
  • #15
I think i was not very clear in putting up my concern. Sorry for that. I am in favor of integration. If you look at the post I quoted above, it is addressed to the guy who meant integration is not necessary to solve this. He was plugging it to ## v = v_o - a*t ## . I think this is wrong. So I asked him to explain himself. I guess I'm clear now.
 
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  • #16
Here is how I would have approached the question
1. Find a reference frame (if possible inertial) such that one of the boats is stationary in it.
2. Use ## a= v*(dv/dx) ## instead of ## a= (dv/dt) ## to directly relate velocity and displacement.
 
  • #17
Vatsal Sanjay said:
Here is how I would have approached the question
1. Find a reference frame (if possible inertial) such that one of the boats is stationary in it.
2. Use ## a= v*(dv/dx) ## instead of ## a= (dv/dt) ## to directly relate velocity and displacement.
You are addressing (b)? kitsh hasn't solved (a) yet.
Using a= v*(dv/dx) would be a good idea if the aim were to find v as a function of x, but (a) asks for v as a function of time.
 
  • #18
haruspex said:
You are addressing (b)? kitsh hasn't solved (a) yet.
In my last comment yes I was. I think I must have mentioned. Ya and for for the (a) part, all someone needs is a little knowledge if basic calculus.
 

Related to When will the fishing boat and the speedboat meet?

1. How does drag force affect the movement of boats?

Drag force is a resistance force that acts in the opposite direction of the boat's movement. This force is caused by the friction between the boat's surface and the water, and it slows down the boat's speed.

2. Can drag force be reduced to make a boat move faster?

Yes, drag force can be reduced to increase the speed of a boat. This can be achieved by making the boat's surface smoother, reducing its size or weight, or using streamlined designs.

3. How do different types of water affect drag force?

The density and viscosity of the water can affect the drag force on a boat. Higher density and viscosity will result in a higher drag force, making it more difficult for the boat to move through the water.

4. What are some ways to calculate drag force on a boat?

There are several methods to calculate drag force on a boat, including the drag coefficient method, the Reynolds number method, and the CFD (Computational Fluid Dynamics) method. Each method uses different equations and factors to determine the drag force.

5. How does the shape of a boat impact drag force?

The shape of a boat can greatly impact the amount of drag force it experiences. A streamlined shape with a pointed front and tapered back will experience less drag compared to a flat or bulky shape. This is because a streamlined shape creates less turbulence and has a smaller surface area for water to push against.

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