relative velocity

**hooke** · February 2nd 2010, 04:36 AM

A canal of width 2a has parallel straight banks and the water flows due north . The points A and B are on opposite banks and B is due east of A , with the point O as the midpoint of AB . THe x-axis and y-axis are taken in the east and north directions respectively with O as the origin . The speed of the current in the canal , $v_c=v_o(1-\frac{x^2}{a^2})$ , where $v_o$ is the speed of the current in the middle of the canal and x the distance eastwards from the middle of the canal . A swimmer swims from A towards the east at speed $v_r$ , relative to the current in the canal . Taking y to denote the distance northwards travelled by the swimmer , show that

$<br /> \frac{dy}{dx}=\frac{v_o}{v_r}(1-\frac{x^2}{a^2})<br />$

**Grandad** · February 2nd 2010, 05:47 AM

Hello hooke

Originally Posted by hooke

A canal of width 2a has parallel straight banks and the water flows due north . The points A and B are on opposite banks and B is due east of A , with the point O as the midpoint of AB . THe x-axis and y-axis are taken in the east and north directions respectively with O as the origin . The speed of the current in the canal , $v_c=v_o(1-\frac{x^2}{a^2})$ , where $v_o$ is the speed of the current in the middle of the canal and x the distance eastwards from the middle of the canal . A swimmer swims from A towards the east at speed $v_r$ , relative to the current in the canal . Taking y to denote the distance northwards travelled by the swimmer , show that

$<br /> \frac{dy}{dx}=\frac{v_o}{v_r}(1-\frac{x^2}{a^2})<br />$

Draw a triangle of velocities such that the velocity of the swimmer relative to the ground = velocity of swimmer relative to the water plus the velocity of the water relative the ground.

This is a right-angled triangle, in which the velocity of the swimmer relative to the ground has a component $v_r$ eastward and $v_c$ northward. So, if $(x,y)$ is his position relative to the ground at time $t$ , we get:

$\frac{dx}{dt}= v_r$

$\Rightarrow \frac{dt}{dx}=\frac{1}{v_r}$

and

$\frac{dy}{dt}=v_c=v_o\left(1-\frac{x^2}{a^2}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$

$=\frac{v_o}{v_r}\left(1-\frac{x^2}{a^2}\right)$

Grandad

**hooke** · February 2nd 2010, 06:29 AM

Originally Posted by Grandad

Hello hookeDraw a triangle of velocities such that the velocity of the swimmer relative to the ground = velocity of swimmer relative to the water plus the velocity of the water relative the ground.

This is a right-angled triangle, in which the velocity of the swimmer relative to the ground has a component $v_r$ eastward and $v_c$ northward. So, if $(x,y)$ is his position relative to the ground at time $t$ , we get:

$\frac{dx}{dt}= v_r$

$\Rightarrow \frac{dt}{dx}=\frac{1}{v_r}$

and

$\frac{dy}{dt}=v_c=v_o\left(1-\frac{x^2}{a^2}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$

$=\frac{v_o}{v_r}\left(1-\frac{x^2}{a^2}\right)$

Grandad

Thanks a lot Grandad , i would like some clarification on some terms used here ,

velocity of swimmer relative to water means his velocity is 'affected' by the velocity of water , so in this case would be the hypothenus of the triangle ?

velocity of swimmer relative to the ground - 'nett' velocity ?

**Grandad** · February 2nd 2010, 08:01 AM

Hello hooke

Originally Posted by hooke

Thanks a lot Grandad , i would like some clarification on some terms used here ,

velocity of swimmer relative to water means his velocity is 'affected' by the velocity of water

Any velocity is made up of (a) magnitude; and (b) direction. For the velocity of the swimmer relative to the water:

(a) the magnitude is his speed in still water; in this question this is $v_r$ .

(b) the direction is the direction in which he is pointing (which is not necessarily the direction in which he's moving, because the current will take him 'off-course'). In this question he's pointing due East.

To get a clearer picture of what this means, imagine that you're in a helicopter flying due North (the direction of the current) directly over the river, moving at exactly the same speed as the water. So the water appears still to you. As you look down, you'll see the swimmer as though he were swimming through still water. That's his velocity relative to the water (and, indeed, relative to you at this moment).

so in this case would be the hypothenus of the triangle ?

No, this is not the hypotenuse of the triangle. The hypotenuse is what you have referred to as his 'nett' velocity - the velocity of the swimmer relative to the ground.

Study the diagram I have attached, and you should see what I mean. If you're still not sure, imagine that the movements of the swimmer and the water take place separately. First the swimmer has a turn, and he swims a certain distance eastwards; then it's the water go, and it carries the swimmer a little way northwards. Well, just allow those two movements to happen at the same time, and you're there.

Grandad

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