# Velocity question

## Homework Statement

A person runs around an equilateral triangle. Each side is 53.3 meters. The person runs completely around the triangle once and 2/3's of the way around the triangle, and the average speed for this trip is 10 m/s. What is the magnitude of the amount of the average velocity for this trip? How do you do this??

## Homework Equations

velocity = distance over time.

## The Attempt at a Solution

the guy ran 53.3 meters x 5 which = a total of 266.5 meters. I divided that by 10 and get 26.65. But thats not the answer. Whats the answer and how do you do it?

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ehild
Homework Helper

## Homework Statement

A person runs around an equilateral triangle. Each side is 53.3 meters. The person runs completely around the triangle once and 2/3's of the way around the triangle, and the average speed for this trip is 10 m/s. What is the magnitude of the amount of the average velocity for this trip? How do you do this??

## Homework Equations

velocity = distance over time.

## The Attempt at a Solution

the guy ran 53.3 meters x 5 which = a total of 266.5 meters. I divided that by 10 and get 26.65. But thats not the answer. Whats the answer and how do you do it?
You divided distance by speed, what did you got? You need to find the magnitude of average velocity.

Velocity is a vector, and it does change along the triangle. What would be the average velocity if the person run once along the triangle?

ehild

• 1 person
that does not help me at all

ehild
Homework Helper
I try to guide you, but you need to show some effort. Check your notes or book. What is velocity?

ehild

• 1 person
Check Again! Your equation for velocity is wrong. That gives speed.
EDIT:Whoops! crossed posts with ehild.

• 1 person
Wait i think i get where you are taking me. the displacement is 53.5 meters. He ran 5 sides of the triangle in 10 seconds. 10 divided by 5 = 2. The answer is 2 because it takes him 2 seconds to get from where he started to where he ended? I know the answer is 2, but is that why. Thank echild

mmm....You are confusing distance with sides now.

well he ran the whole triangle once which is 3 sides and then he ran 2 more sides which makes 5 total. his total displacement doesnt matter though am i correct? His displacement is just 1 side. and he runs each side in 2 seconds because 10 seconds divided by 5 sides is 2. i know the answer is 2 but is my reasoning correct?

What is the magnitude of the amount of the average velocity for this trip? How do you do this??
What are you supposed to find time or avg. velocity?

. his displacement doesnt matter though am i correct?
Nope, you are not, not when you are supposed to find velocity. You have either confused definition of displacement or that of velocity.

His displacement is just 1 side.
What is one side?

"What is the magnitude of the amount of the average velocity for this trip?" I have to find avg velocity. I have the answer on my study guide, the answer is 2. I just want to know how to get to 2 so i can perform it on the test

ehild
Homework Helper
Wait i think i get where you are taking me. the displacement is 53.5 meters. He ran 5 sides of the triangle in 10 seconds. 10 divided by 5 = 2. The answer is 2 because it takes him 2 seconds to get from where he started to where he ended? I know the answer is 2, but is that why. Thank echild
Yes, the average velocity has magnitude 2m/s. But it did not take 2 seconds to reach to the final position. If L is the length of one side of the triangle, the person covers one side in t=L/v(speed) =53.5/10=5.35 s.
Running along five sides of length L each, the person is one side away from its initial position. The magnitude of the displacement is L and the distance covered (the length of the whole path) is 5L. Velocity is displacement over time, velocity=L/t; speed is distance covered over time speed=5L/t.

ehild

Last edited:
Chestermiller
Mentor
In your first post, you calculated the number 26.65. This is actually the total number of seconds it took for him to run the 5 sides of the triangle (rather than a velocity or speed). In another posting, you determined that his total displacement was 53.3 meters. Average velocity is defined as the total displacement divided by the total time.