# Relative Velocity - I know the answer - But how to arrive at it?

1. Mar 4, 2006

### nyclio

I have been pulling my hair out over this one. Here is the question:

Assume that a river has straight and parallel banks and that the current is 0.75 m/s. Drifiting down the river, you fall out of your boat and immediately grab on to a log. You hold on for 40 seconds, and then swim after the boat with a speed relative to the water of 0.95 m/s. The distance of the boat downstream from the log when you catch it is:

ANS: 54 meters.

How did they arrive at this answer? (no work is shown) I thought it would be 40*.75 (since the boat is moving - i don't see what it has to do with your speed of .95 since that would only effect time taken to reach the boat)

Any help here is appreciated.

2. Mar 4, 2006

### Lyuokdea

The best way to approach a problem like this in my opinion is to divide it into two parts:

Part I: The motion of the boat while you are at rest (with respect to the boat) - You want to find the initial condition of the boat when you start swimming after it. When you begin swimming after the boat, how far away is the boat from you? Let's call this initial distance D.

Part II: How far does the boat travel while you catch up with it? There are two subparts here:

Subpart A: How much time passes between the time you begin swimming and the time when you catch up with the boat? To solve this part. In this part you are right in assuming that the speed of the current doesn't matter, because the boat has the same speed as the speed of the current, and you can simplify the problem by ignoring that term for the moment. You need two pieces of information, how far the boat is away from you, and how fast you are moving towards it. We'll call this time T.

Subpart B: How far does the boat travel in time T? Now you have to deal with the current again. During the time you were catching up with the boat, the boat is continuing to move forward at a constant velocity equal to that of the current. We'll call this distance x.

Now, to find the total distance that the boat travelled, so simply add up your two distances from Part I, and Part 2 Subpart B, and that is your final answer.

~Lyuokdea

3. Mar 4, 2006

### nyclio

Thank you! I finally got it!

4. Mar 4, 2006

### sebas531

i dont really understand the question of the problem. After analyzing and working out the problem myself i have come to certain answer for what i believed the problem asked; however, according to the answer given, 54m/s, which i came across in the problem, i can make a conclusion that the problem asked : WHAT IS THE DISTANCE OF THE CANOE FROM THE TREE, AT THE TIME THAT ONE REACHES THE EXACT POSITION THAT THE CANOE TRAVELED WHILE ONE WAS ATTACHED TO THE TREE, 30m, is that what the question was really asking?

5. Mar 4, 2006

### nyclio

well a boat is moving along at .75 m/s, you fall out and grab this log...you hold on to it for 40 seconds...the boat continues to move in the mean time...then you start swimming (at a speed of .95 m/s) from the log to go catch the boat...the question wants to know - by the time u actually catch the boat...how far is it from the log u had grabbed on to...and here's how u get 54:

for the 40 seconds u are holding on to the log, the boat has travelled 30 meters (40*.75)

then u start swimming to catch it...how long does it take u to swim 30 meters at .95 m/s?

time = dist. / vel.
= 30 / .95
= 31.57 seconds.

Now you are swimming for 31.57 seconds before you actually catch the boat. However, the boat was moving during that time too.

31.57s * .75 (boat vel.) = 23.68

Add that to the original distance of 30 for 53.68...or 54.

6. Mar 5, 2006

### sebas531

The swimmer does not catch the boat 54m from the log. That 23.68 is how far the boat moves from the 30meter "point" while the swimmer was swimming towards that 30 meter "point," but it does not mean that the swimmer actually cought up to the boat.

I believe that the swimmer catches up to the boat 142.5 meters away from the the tree and i will explain how i got that.

We know that at a certain time the swimmer is going to catch up to the boat, so lets make an equation in which that time will be represented by the variable x. .95(m/s)x=.75(m/s)x+30(m) once again x represents the time that it will take the swimmer to catch the boat. I used +30 because the boat is 30 meters away from the swimmer.
then we get x= 150(s) this is the time that it will take for the swimmer to catch up to the boat.
We know the velocity of the swimmer from the tree,.95m/s, by using v=d/t we get d=vt

which is d=(.95m/s)(150s)=142.5m

7. Mar 5, 2006

### GregA

Not sure I agree with that...we know that the velocity of the swimmer relative to the current is .95m/s and because the current is affecting the motion of the boat then...

the swimmer's velocity relative to the boat is .95m/s

If we consider all motion relative to the boat, then the boat is stationary (for now) and a swimmer travelling at .95m/s is travelling towards it from a distance of 30m away. and it takes him (30/.95) seconds to reach the boat.
If we now assume that the log was stationery at all of this time and unaffected by the current (it must be to fit the answer) then the speed at which the swimmer was travelling relative to the log is .95 + .75 m/s (1.7m/s) and the swimmer maintains this speed for (30/.95) seconds travelling just under 54m

If the log *was* to some extent being moved by the current then it would be even closer to the boat not further away

just my 2 pence worth

8. Mar 5, 2006

### sebas531

Why would one add the distance that the boat moved in the time that it took the swimmer to get to the 30meter mark? The boat moved, it doesnt mean the swimmer caught it.

so the current of the river is making the swimmer go faster? if that is the case then i guess everything i did is wrong since i thought that the speed man counting the current speed was .95......ok......now if it is also in that way that the current was making the swimmer go faster how come you didnt use that speed when calculating how long it took for the swimmer to get to the 30 meter mark?

I think that i might not be understanding the problem througly, and i just want to understand it.

9. Mar 5, 2006

### GregA

well the swimmer is travelling at (.75 + .95)m/s...the boat however is drifting along with the current at a speed of .75m/s
considering all motion relative to the log then the swimmer is travelling at a speed of .95m/s (imagine if they were both just drifting at .75m/s (ie: the swimmer is just letting himself be carried by the river)...from the log or swimmers point of view neither would be moving because the distance between them is constant)

It is given however that from the water's point of view (which in turn directly influences the boats point of view) the swimmer has a speed of .95m/s
"and then swim after the boat with a speed relative to the water of 0.95 m/s."
The time taken to reach the boat involves the distance the swimmer is away from the boat and his relative speed...since the log is stationery (else it wouldn't be 54m away) relaitive to the log then the swimmer has been displaced by his true speed *time (1.7(30/.95)) metres

Last edited: Mar 5, 2006
10. Mar 5, 2006

### sebas531

Im still not quite understanding you,GregA;however, what do you have to say about the method that the other people used? while i was working the problem i did come across that 54m, but i did not take it as my answer b/c i, for some reason am not able to "see" how it would work.

11. Mar 5, 2006

### GregA

I think the other methods are perfectly acceptable...to put it in my terms..

the swimmer is holding on the log whilst the boat drifts away at a speed of .7m/s (don't know why he did this )...but then he decides that the water is too cold and figures he should now try and catch the boat again...as soon as he lets go of the log the water starts to pull him at the same speed as the boat (cos someone tied an anchor to the log ) and then he swims .95m/s faster so that he can catch up with it.

adding .75t to .95t is a shortcut to simply 1.7t (whilst .95m/s is used to calculate how long it takes to catch up with the boat...the fact that they are travelling an additional .75m/s isn't taken into consideration until it is time to figure out how far the swimmer has been displaced relative to the log (not the boat))

Last edited: Mar 5, 2006
12. Mar 5, 2006

### sebas531

GregA, thanks for trying to explain it, i do like the story you made just now. Anyways, i was shaving my face and i was thinking about the problem, and i get it. The problem i really had was that i thought the man was traveling at .95 with the current, but i get it. I guess i should go and shave my face everytime i have problem, huh?. Thanks though