# Dog swimming

1. Dec 1, 2008

### Throwback24

1. The problem statement, all variables and given/known data

A dog wishes to swim across a slow-moving stream. The dog can swim at 2 m/s in calm water. The current velocity is 3 m/s. The distance directly across the stream is 50 m.

If the dog points himself directly across the stream how long will it take to get across the stream?

How far downstream will the current have carried the dog when the dog gets to the other side?

What was the dogs velocity relative to the bank from where the dog started?

2. Relevant equations

3. The attempt at a solution

With a.) I'm given how fast the dog can swim and velocity of the stream? I need to find T.

b.) I don't know what they're asking :(

c.) I don't now what they're asking :(

Can you guys please break it down for me? I think with B you multiply velocity by time.

2. Dec 1, 2008

### horatio89

a) Velocity = displacement/time, therefore, time = ?. To find velocity, simply add the velocity of the stream and the dog. (Note: Recall that velocity is a vector). For this part, we are interested in the component of velocity perpendicular to the bank.

b) Well, from a), you'll realise that the dog doesn't actually cross the stream perpendicular to the bank because the current pushes it downstream, but at an angle. Find its displacement in the direction parallel to the bank, which can be found using the time in part a) and the component of velocity parallel to the bank.

c) This is the velocity found in a). (Note: Remember once again, velocity is a vector. There must be 2 pieces of info provided)

Last edited: Dec 1, 2008
3. Dec 1, 2008

### Throwback24

Time=Displacement/Velocity?!

How do you know the dog is it an angle? I thought the dog was facing the stream

4. Dec 2, 2008

### horatio89

Yes, but imagine this, the dog tries to swim at a velocity (and hence a route) perpendicular to the bank, but the stream has a velocity parallel to the bank, pushing it downstream as it swims. Wouldn't it be moving at an angle?

I'm sorry if I wasn't clear earlier, for a), your equation is correct... We are concerned with the component of displacement perpendicular to the bank (i.e. the distance from bank to bank) and the component of velocity perpendicular to the bank, which is solely contributed by the dog's swimming velocity.

5. Dec 3, 2008

### Throwback24

The component of the displacement? I'm so lost lol... can you run that by me again please?

Is it like this?

------>--------->
2 m/s 3 m/s

Those are the two vectors no?

6. Dec 9, 2008

### Throwback24

A.

T=D/S
=50 m/2m/s
=25 seconds

B.

C2=A2+B2

Is that correct? It's a right angle triangle.

..............
.
.
.
.
.

Up is 2 m/s and sideways is 3 m/s. The resultant would be the hypotenuse correct?

How would I go about getting C?