Calculating Velocity and Its Signs

  • Thread starter mxh91
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  • #1
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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!
 

Answers and Replies

  • #2
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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.
 
  • #3
K^2
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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:
  • #4
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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?
 
  • #5
jtbell
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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.
 
  • #6
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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)
 
  • #7
K^2
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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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