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Speed vs velocity

  1. Sep 19, 2004 #1
    I'm getting confused with speed and velocity. I know that speed does not tell direction and velocity does but here's the question:

    A person walks first at a constant speed of 5.00m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00m/s.

    What is the average speed and average velocity?

    I pretended that point A to B is 15m and calculated that the average speed would be 3.75s
    Is there another way to calculate the speed if I had not made up a distance from A to B?

    Also, what would the velocity be? Would it be the same as the speed? Since the velocity would be +5.00m/s for the first part then -3.00m/s how do I calculate the velocity?
     
  2. jcsd
  3. Sep 19, 2004 #2
    As far as I know, velocity is speed with a direction.

    I'm not sure on the rest though, sorry.
     
  4. Sep 19, 2004 #3

    HallsofIvy

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    It is impossible to answer the first question without knowing the distance from A to B (or, same thing, the time required). Yes, if the distance was 15 m (good choice: its divisible by both 3 and 5: It took 15/5= 3 seconds to go from A to B and 15/3= 5 seconds to go from B to A. Total distance 30 m, time 8 sec so average speed is 30/8= 15/4= 3.75 m/s exactly what you said (except you had 3.75 s!).

    On the other hand, if the distance from A to B were 30 m (still divisible by both 3 and 5!) the time to go from A to B would be 30/5= 6 s and the time to go from B to
    A would be 30/3= 9 s. Now, he has gone 60m in 15 s. : average speed 60/15= 4 m/s.]

    It is impossible to determine the average speed without knowing the distance.

    Answering the second question is easy! The person's DISPLACEMENT (distance from initial point) is 0 because he returned to his initial point. No matter what the time, t, required his average VELOCITY is 0/t= 0.
     
  5. Sep 19, 2004 #4
    Yeah, velocity is a vector so has magnitude and direction where as speed is a scalar and has magnitude only.

    Speed = Distance/Time

    Velocity = Displacement/Time
     
  6. Sep 19, 2004 #5
    what is the difference between the average velocity and velocity?

    If the person walks first at a constant speed of 5.00m/s along a straight line from point A to point B and then back along the line from B to a point before point A at a constant speed of 3.00m/s, what would the difference between average velocity and velocity be?
     
  7. Sep 19, 2004 #6

    Pyrrhus

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    Halls, i think he can do it.

    Ok so let's do this:

    info
    He covered a distance [tex] x [/tex]

    [tex] V_{1} = 5 m/s [/tex]

    [tex] V_{2} = 3 m/s [/tex]

    Using this equation

    [tex] V = \frac{d}{t} [/tex]

    [tex] t_{2} = \frac{x}{V_{2}} [/tex]

    [tex] t_{1} = \frac{x}{V_{1}} [/tex]

    Remember Average Speed is Total Distance divided by Total Time so

    [tex] t_{1} + t_{2} = T_{total} [/tex]

    [tex] 2x = D_{total} [/tex]

    so

    [tex] V_{ave} = \frac{2x}{\frac{x}{V_{2}} + \frac{x}{V_{1}}} [/tex]

    solving it

    [tex] V_{ave} = 3.75 m/s [/tex]
     
  8. Sep 19, 2004 #7

    Pyrrhus

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    Velocity or Instant Velocity, it's just dv/dt, which means a instant value, which could be different before the body reached it or after it reached it, it's like a velocity at x point.

    Average velocity on the other hand it's displacement divided by time change of the whole movement, from where it started to where it finished. In your case the body started a 0 so the displacement vector went from that and then returned to the exact point, so the displacement became 0.

    Velocity is a vector while Speed is a Scalar.
     
  9. Sep 19, 2004 #8
    What if I started 5 meters away from the origin and came towards the origin at 5m/s for 2 seconds. Then what would the velocity be? would it be -5m/s?

    However, my displacement is 5m so would it be -2.5m/s?
     
  10. Sep 19, 2004 #9

    Pyrrhus

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    Your Average Velocity will be 0, still. Your Displacement was 0.

    You covered a 5 meter a amount, and the you covered the same.

    it will be like 0-0 =0

    Remember

    [tex] \Delta x = x - x_{o} [/tex]

    So you initial X will be 0 and your final X will be 0 too!
     
    Last edited: Sep 19, 2004
  11. Sep 19, 2004 #10

    Pyrrhus

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    Remember displacement only sees final and initial, if you came from the origin at any speed for any time and came back to the origin again, your average velocity will still be 0!
     
  12. Sep 19, 2004 #11

    Pyrrhus

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    Oh i misread :rofl:, yes. (to the -2.5 m/s)
     
    Last edited: Sep 19, 2004
  13. Sep 19, 2004 #12
    I understand that but re-read the question, I came back towards the origin for 2 seconds at 5m/s, that means I have gone past the origin which is only 5m away. Therefore, I have a displacement of -5m. So, what would the velocity be? would it be -5m/s?

    Since my displacement is -5m, would my velocity ve -2.5m/s?
     
  14. Sep 19, 2004 #13

    Pyrrhus

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    read above i noticed later
    :biggrin:
     
  15. Sep 19, 2004 #14
    Okay, still confused :P

    Walking away from the origin gives a positive velocity. Walking towards the origin gives a negative velocity. What if I was behind the origin, and walking towards it? is it still positive? What if I was behind the origin and walked away? That's walking away from the origin, so that means it's positive? so would the velocity to my lasy question be positive 2.5m/s?
     
  16. Sep 19, 2004 #15

    Pyrrhus

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    It's just a sign convention you pick.. The standard is right and up positive and down and left negative.

    If you walk towards the origin the vector is pointing left, so it's negative under that convention. It will be positive under that convention wallking toward the origin from behind (i assume you mean down), it will be negative because of that convention walking away from the origin if you were behind. It will be negative.
     
  17. Sep 19, 2004 #16
    What if I walked away from the origin 10m to the right for 2s, then walked back 5m in 3 seconds. My total displacement is 5m, my time is 5s so my velocity is 1m/s?
     
  18. Sep 19, 2004 #17

    Pyrrhus

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    Remember is displacement divided by time change, so only 1 second.
     
  19. Sep 19, 2004 #18
    but it took me 5s to get there, how is it 1s?
     
  20. Sep 19, 2004 #19

    Pyrrhus

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    Well by applying the definition of Average Velocity.

    Thinking it logically, you walked for 3 seconds to a x place then in the return it took you 2 seconds, so logically you might think 3+2=5, will be total seconds taken, but Average Velocity actually means a slope of a triangle with opposite side displacement and a adjacent side of change of t. The problem i see with you, is you are interpreting velocity the same as speed, in physics they do not mean the same!.
     
  21. Sep 19, 2004 #20
    OKay, I'm just making up random questions so I can understand this concept.

    If I started 10m away from the origin, I walked 5m towards it in 1s. Would my velocity be negative 5m/s or positive?

    Lets say I started 10m away from the origin, and walked to it in 2s. My velocity is zero?? Would it be -5m/s since I STARTED away from the origin?

    I tested this on a motion sensor...I walked towards it at a constant speed and it showed a negative velocity. However, if I'm returning to the origin, wouldn't the velocity be zero?
     
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