Fisherman's boat Relative motion

[heythere1010]

Homework Statement

A fisherman wishes to cross a river to the shore directly south of his position. The river is 400 m across, and the fisherman's boat is capable of a speed of 2.5m/s. If the river's current has a velocity of 1.5m/s [W], determine

a) the time it takes the fisherman to cross the river (200 s)
how far downstream he will be before he reaches the other shore (300 m)

This is the part I didn't answer.

If the fisherman wished to land on the shore directly south of his position, how long would it him take to cross the river, and what would be the ground velocity of the boat?

Homework Equations

d =v*t

The Attempt at a Solution

I have calculated the time from question 1, but I'm not sure what is different about time in question 2. I was thinking the ground velocity is 2 m/s since it would be d/t.

 

[jbriggs444]

How did you calculate the 200 second answer for a)?

Edit: Scratch that. I see how you got 200 seconds. How did you get 300 meters?

 

[heythere1010]

How did you calculate the 200 second answer for a)?

Edit: Scratch that. I see how you got 200 seconds. How did you get 300 meters?

I made a triangle with 2.5m/s as the hypotenuse, 1.5m/s as one of the sides. and solved for the other side. I did t=d/v (400/2=200).

 

[jbriggs444]

Right. Now how did you get 300 meters?

 

[heythere1010]

Right. Now how did you get 300 meters?

1.5(200)

 

[jbriggs444]

And what is 1.5?

 

[heythere1010]

And what is 1.5?

Velocity of the current.

 

[jbriggs444]

Is it the same as the downstream velocity of the boat?

 

[heythere1010]

Is it the same as the downstream velocity of the boat?

No.

 

[jbriggs444]

The question asks how far downstream the boat goes, not how far downstream the current goes. So multiplying by 1.5 would not be appropriate. But let's back up and re-examine what the question is asking...

Homework Statement

A fisherman wishes to cross a river to the shore directly south of his position. The river is 400 m across, and the fisherman's boat is capable of a speed of 2.5m/s. If the river's current has a velocity of 1.5m/s [W], determine

a) the time it takes the fisherman to cross the river
how far downstream he will be before he reaches the other shore

If the fisherman wished to land on the shore directly south of his position, how long would it him take to cross the river, and what would be the ground velocity of the boat?

As I read this, there are two parts with two questions each.

The first part (which you answered) assumes that the fisherman aims his boat due south relative to the moving water. So it drifts downstream at the same rate as the current. The second part (about which you are asking) assumes that the fisherman aims his boat at an angle so that his boat does not drift downstream.

If this is the case, what is the boat's southward velocity for the first part?

 

[heythere1010]

The question asks how far downstream the boat goes, not how far downstream the current goes. So multiplying by 1.5 would not be appropriate. But let's back up and re-examine what the question is asking...As I read this, there are two parts with two questions each.

The first part (which you answered) assumes that the fisherman aims his boat due south relative to the moving water. So it drifts downstream at the same rate as the current. The second part (about which you are asking) assumes that the fisherman aims his boat at an angle so that his boat does not drift downstream.

If this is the case, what is the boat's southward velocity for the first part?

Yes. I calculated the angle in another part of the question. I got 53 degrees.

 

[jbriggs444]

Are you assuming that the boat's ground speed is 2.5 meters per second or that the boat's speed with respect to the water is 2.5 meters per second? The problem assumes the latter.

53 degrees is not consistent with the intended interpretation of the problem.

 

[heythere1010]

Are you assuming that the boat's ground speed is 2.5 meters per second or that the boat's speed with respect to the water is 2.5 meters per second? The problem assumes the latter.

53 degrees is not consistent with the intended interpretation of the problem.

The boat's speed is 2.5m/s

 

[Nathanael]

The first part (which you answered) assumes that the fisherman aims his boat due south relative to the moving water.

That's not how I understand it. It says, "A fisherman wishes to cross a river to the shore directly south of his position. I understand this to mean he lands (not aims) directly south.

a) the time it takes the fisherman to cross the river (200 s)

Did you get this time assuming he aims south? Can you show how you got the time?

I was thinking the ground velocity is 2 m/s since it would be d/t.

That's one way to do it, since it's constant. I'm not exactly sure what you did to get the time. I found the time from the relative velocity. And this is the time I get for landing south, not aiming south.

 

[jbriggs444]

That's not how I understand it. It says, "A fisherman wishes to cross a river to the shore directly south of his position. I understand this to mean he lands (not aims) directly south.

That's how I read it at first as well. But the existence of the second part of the question makes that interpretation untenable.

 

[Nathanael]

That's how I read it at first as well. But the existence of the second part of the question makes that interpretation untenable.

Or it could be a simple conceptual question not intended to be calculated.

 

[jbriggs444]

Or it could be a simple conceptual question not intended to be calculated.

The numbers are too carefully chosen for that interpretation to be tenable.

 

[heythere1010]

I made a triangle with 2.5m/s as the hypotenuse, 1.5m/s as one of the sides. and solved for the other side. I did t=d/v (400/2=200). That's how I solved time.

 

[Nathanael]

The numbers are too carefully chosen for that interpretation to be tenable.

Why's that? The speed of the boat is more than that of the river.

 

[jbriggs444]

And that's the wrong approach for the first part. 2.5 m/s is not the hypotenuse.

 

[Nathanael]

I made a triangle with 2.5m/s as the hypotenuse, 1.5m/s as one of the sides. and solved for the other side. I did t=d/v (400/2=200). That's how I solved time.

Why is 2.5 the hypotenuse? 2.5 m/s is the speed of the boat relative to water, not relative to the shore.

 

[jbriggs444]

It's a 3-4-5 triangle. That's an unlikely coincidence.

 

[Nathanael]

It's a 3-4-5 triangle. That's an unlikely coincidence.

No coincidence necessary; it's not a real fisherman. Are you saying it's impossible for the boat to reach the shore directly south of where it started?

 

[jbriggs444]

No coincidence necessary; it's not a real fisherman. Are you it's impossible for the boat to reach the shore directly south of where it started?

Certainly it's possible. But that's not what part 1 is asking for. Part 2 asks about reaching the point on the shore due south. And that's where the original 3-4-5 triangle comes in.

 

[heythere1010]

Why is 2.5 the hypotenuse? 2.5 m/s is the speed of the boat relative to water, not relative to the shore.

So do I just do 400/2.5(his speed)?

 

[Nathanael]

So do I just do 400/2.5(his speed)?

To find the time aiming south, yes.

 

[jbriggs444]

Yes. It is just that simple. His southward velocity is 2.5 meters per second and he needs to cover 400 meters southward.

 

[heythere1010]

How about for the second part of finding the time?

 

[jbriggs444]

How about for the second part of finding the time?

Now you can draw that right triangle. But label the sides to keep track of what each one represents.

First, the southward pointing side. This is the southward component of the boat's velocity relative to the water.

Next the eastward (or westward) pointing side. This is the upstream component of the boat's velocity relative to the water. This must be 1.5 meters per second in order to avoid driving upstream or drifting downstream.

Finally the hypotenuse. This is the boat's total velocity relative to the water. Its length is given by the max speed of the boat, 2.5 meters per second.

You had already solved this triangle for the southward velocity represented by the unknown side.

 

[RENI2345]

so this is how I did it, I think it makes sense