# Solving Mechanics Problem: Velocity & Distance of Swimmer B

• Spoony
In summary, Swimmer A swims due east at a constant speed of 3m/s along a straight stretch of river with neglible current, keeping a constant distance of 10m from the southern bank. A second swimmer, B, starts swimming from this bank when A is a disnace L down the river from her (i.e. L is the distance measured along the bank.) Swimmer B swims with a constant speed of 2m/s and at an angle of N60E in order to intercept A.Va = 3i m/sVb = sqrt(3)i + 1jVa = (3-sqrt(3))i + 1jB has
Spoony

## Homework Statement

Swimmer A swims due east at a constant speed of 3m/s along a straight stretch of river with neglible current, keeping a constant distance of 10m from the southern bank. A second swimmer, B, starts swimming from this bank when A is a disnace L down the river from her (i.e. L is the distance measured along the bank.) Swimmer B swims with a constant speed of 2m/s and at an angle of N60E in order to intercept A.

i) what is the velocity of b as observed from a?
ii) what is the distance L, and how long does B swim for?
iii) If a strong current is flowing how, if at all, would the above results be alterd?

## The Attempt at a Solution

I have calculated A and B's vector velocities.
Va = 3i m/s
Vb = sqrt(3)i + 1j
Then surely the velocity of b as observed from is just the difference between the two vectors.
Vdiff. = (3-sqrt(3))i + 1j

Then i run into trouble for the next bit, surely if swimmer b's vertical component of velocity is 1m/s then no matter what distance A swims before B starts. B will always have to swim through 10m of water vertically. So if B's vertical component of velocity is 1, then he has to swim 10s before he contacts A.
Its the distance L which confuses me. B's horizontal component is sqrt(3) which is less than 3, so if B starts after A has passed B will never reach A. The distance will get progressively larger. So surely B has to start before A passes him on the river?

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Spoony said:
I have calculated A and B's vector velocities.
Va = 3i m/s
Vb = sqrt(3)i + 1j
Recheck the velocity of B. [Edit: Your velocity is fine!]
Its the distance L which confuses me. B's horizontal component is sqrt(3) which is less than 3, so if B starts after A has passed B will never reach A. The distance will get progressively larger. So surely B has to start before A passes him on the river?
Right. A is a distance L down river from B. When B starts moving, A is to the west of B. (As you realize, that's the only interpretation that makes sense.)

[Edit: Ignore my comment about velocity.]

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The velocity of B is right (im sure), as the triangle is a 30,60,90 degree triangle, since 2 is the hypotneuse then root3 and 1 are the other sides. The angle used is 30, as N60E is 60 degrees FROM the north so its 30 degrees measured from the bank counterclockwise.

Correct me if I am wrong. But I am preety sure that's right. as the triangle is of the form:

http://id.mind.net/~zona/mmts/miscellaneousMath/tri454590306090/t306090.gif

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Ahhh, going mad. I can't see where I am going wrong as the vertical component from the triangle is 1m/s and the horizontal is sqrt(3)m/s. Unless I've got my i's and j's mixed up. Please help me :(

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D'oh! I just realized that I'm full of it. Your Vb components are correct. (Sorry about that!)

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Hehe np, would the time taken be 10s and the distance L be 15.58m then?
(as |10*sqrt(3) - 3*10| = 15.58 ?)
sorry to badger, but I am quite unconfident in my physics atm as I am having to learn this all by myself as I've missed a lot of lectures.

Spoony said:
Hehe np, would the time taken be 10s and the distance L be 15.58m then?
(as |10*sqrt(3) - 3*10| = 15.58 ?)
The time is 10s, but check your arithmetic on that subtraction.

Sorry 12.68m :S

Looks good.

Cheers for all your help mate :)

## 1. What is the formula for calculating velocity?

The formula for calculating velocity is: velocity = distance/time.

## 2. How do you determine the distance traveled by Swimmer B?

To determine the distance traveled by Swimmer B, you can use the formula: distance = velocity x time.

## 3. Can you explain the concept of velocity in relation to swimming?

Velocity in swimming refers to the speed at which Swimmer B is moving in a particular direction. It is calculated by dividing the distance traveled by the time taken to cover that distance.

## 4. How does the velocity of Swimmer B affect their overall performance?

The velocity of Swimmer B is a crucial factor in their overall performance. A higher velocity means they are able to cover more distance in a shorter amount of time, resulting in a better performance. It also indicates their strength and efficiency in the water.

## 5. Can you provide an example of a mechanics problem involving velocity and distance of a swimmer?

Sure, here's an example: Swimmer B starts at one end of a 50-meter pool and swims at a velocity of 2 meters per second. How long will it take them to reach the other end of the pool? To solve this problem, we can use the formula: time = distance/velocity. In this case, the distance is 50 meters and the velocity is 2 meters per second, so the time taken to reach the other end of the pool would be 25 seconds.

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