Calculating Velocity and Its Signs

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Discussion Overview

The discussion revolves around the concepts of velocity and average speed, particularly focusing on how to calculate them based on displacement and distance traveled. Participants explore examples involving movement in different directions and the implications of these movements on velocity and average speed calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Michael introduces the concept of velocity and presents examples to clarify his understanding, calculating displacement and velocity based on his movements.
  • Some participants emphasize that velocity includes a directional component, contrasting it with speed, which does not have a sign.
  • There is a discussion about average speed, with some participants asserting it is calculated using total distance traveled divided by time, while others clarify the distinction between average velocity and average speed.
  • One participant explains that the average speed in the example provided is 35 km/h, based on the total distance of 70 km over 2 hours, while the average velocity is 15 km/h or -15 km/h depending on the direction of travel.
  • Another participant introduces the concept of triangle inequality to explain the relationship between average speed and average velocity, noting that average speed can be greater than the absolute value of average velocity when direction changes occur.

Areas of Agreement / Disagreement

Participants generally agree on the definitions of average speed and average velocity, but there is some contention regarding the calculations and implications of these concepts in specific examples. The discussion remains unresolved on certain aspects, particularly regarding the clarity of instantaneous velocity in the provided scenarios.

Contextual Notes

Some participants express confusion about the calculations of average speed and velocity, indicating a need for more detailed examples or explanations. The discussion highlights the importance of distinguishing between distance traveled and displacement in these calculations.

mxh91
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Hi, I'm Michael. I am new to Physics. I just learned about velocities from an online website and I still don't quite get it. I know that in order to find velocity, I have to take the the displacement and divide it by time:
V = (delta)d / (delta) t

Here's what I am confused about. Let's say I walk 50 km East, and then 20 km West in 2 hours. My displacement would be +30 km because of final - initial (50 - 20 = +30). Thus, my velocity would be 15 km/h.

Now let's say I walk 20 km East, and then 50 km West in 2 hours. My displacement would be -30 km because of final - initial (20 - 30 = -30). Would my velocity be -15 km/h?

Thanks!
 
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Yes, a key to the idea of velocity is that it has a direction. You can say I traveled -15 km/hr or you can say I traveled 15 km/hr in the west direction. However you say it, a direction has to be included or implied.

If you were talking about speed you would just say 15 km/hr with no negative sign and no direction stated.
 
Average speed here is 35km/h. In general [itex]\| Avg(V)\|\leq Avg(\|V\|)[/itex]

[Having problems with /bar and /overline for some reason, oh well.]
 
Last edited:
K^2 said:
Average speed here is 35km/h. In general [itex]\| Avg(V)\|\leq Avg(\|V\|)[/itex]

[Having problems with /bar and /overline for some reason, oh well.]

Sorry, can you show me how the average speed is 35 km/h?
 
mxh91 said:
Let's say I walk 50 km East, and then 20 km West in 2 hours. My displacement would be +30 km because of final - initial (50 - 20 = +30). Thus, my velocity would be 15 km/h.

Now let's say I walk 20 km East, and then 50 km West in 2 hours. My displacement would be -30 km because of final - initial (20 - 30 = -30). Would my velocity be -15 km/h?

You are correct in both cases, except that I would say that your average velocity is 15 km/h in the first case and -15 km/h in the second case.

In general you need to distinguish between average velocity and instantaneous velocity. In your examples, it's impossible to say what the instantaneous velocities are, because you didn't give enough detail.
 
mxh91 said:
Sorry, can you show me how the average speed is 35 km/h?

Average speed is distance traveled divided by time. (rather than displacement divided by time)
 
mxh91 said:
Sorry, can you show me how the average speed is 35 km/h?
Average speed is distance traveled divided by time. (rather than displacement divided by time)
To answer both of these, no. Average speed is length of path divided by time. While displacement is only 30km, the entire path is 70km. The average speed is therefore 35km/h.

The traveler needs to travel the first 50km before turning around and going 20km back. That can't be done going 15kph. That has to be done going 35kph - on average.

This is an example of triangle inequality. Absolute value of the sum is less than or equal to sum of absolute values. Same for averages. Absolute value of an average is less than or equal to average of absolute values. So absolute value of average velocity is less than or equal to average speed. The "less than" happens when the values change sign/direction. Here, we have velocity changing sign, and so the average speed comes out significantly greater than absolute value of average velocity.
 

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