# Swimmer's velocity relative to the shore (vectors)

ulfy01

## Homework Statement

A swimmer is training in a river. The current flows at 1.33 metres per second and the swimmer's speed is 2.86 metres per second relative to the water. What is the swimmer's speed relative to the shore when swimming upstream? What about downstream?

Pythagoras.

## The Attempt at a Solution

Here's my problem. Because we're looking at vectors, I would normally do the vector sum of both the velocities and use Pythagoras, as both given vectors are perpendicular.

Vcurrent = 1.33 m/s
Vswimmer relative to water = 2.86 m/s

So the resultant vector would be: $\sqrt{}$(1.332 + 2.862)

Giving 3.15 m/s, however this is wrong, as the answer given is 1.53 m/s upstream.

I'm puzzled as to how this answer was reached.

## Answers and Replies

azizlwl
The swimmer is not swimming across the river.
He is swimming against the current. That how normally swimmers trainned.
Now imagine you running in direction to the east at a speed 1.33 metres per second on a train with 1.33 metres per second speed heading west.
To the man on the platform seeing you not moving, but with respect of the train you are running at 1.33 metres per second in easterly direction.

Homework Helper
Gold Member
Hi, ulfy01.

Note that "swimming upstream" means swimming in a direction opposite to the current, not perpendicular to the current. So, there is no right triangle here.

[oops: I'm a bit late here, sorry.]

ulfy01
I just realized this and made a fool of myself, really. Way to overthink a problem and not read it properly. I'll go hide in a corner now. Thanks azizlwl!