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Is my textbook wrong about velocity/displacement vectors?

  1. Feb 1, 2017 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    My textbook states that "The velocity has the same direction as the displacement".

    I feel this statement is incorrect. Keeping in one dimension, let's say that I move in the +ve direction at velocity v for some time t. My displacement is vt which points in the +ve direction. However I then stop and move back to my starting point at velocity -v for time t1 < t. My displacement is now vt - vt1, which still points in the +ve direction, however my velocity is -ve.

    Is my reasoning correct? I don't know if the book is trying to mean something else, or what.

    Thank you!

    EDIT: I believe the book may be talking about average velocity and not instantaneous velocity. So perhaps it's correct in that sense???
     
  2. jcsd
  3. Feb 1, 2017 #2
    The textbook is correct.
    In your example, you state that your displacement is vt - vt1. But displacement is your final position minus your initial position. In this case your final position is vt1 and your initial position is vt. Thus your displacement vt - vt1 but is vt1 - vt which is negative.

    The rate of change of a vector will always point in the same direction as the change in a vector. Otherwise the formula ∆x = vt would not hold because the directions would not coincide. This is true for position and velocity, where the displacement is the change in position, and the velocity is the rate of change of displacement. This is also so true for velocity and acceleration: the change in your velocity will always point in the same direction as the acceleration.

    Note that it is only the changes that point in the same direction as the rate of change. The position does not need to point in the same direction as the velocity. For instance, if you move towards something, your velocity is positive, but your position is negative. The same applies for acceleration and velocity: you slow down when your acceleration is opposite your velocity.
     
  4. Feb 1, 2017 #3
    But my initial displacement is 0 because that's where I started from isn't it? I have attached a crude picture to help explain myself: http://imgur.com/a/LuvLr

    This is why I'm now assuming it's talking about the average velocity since the formula is vavg = ∆x / t
     
  5. Feb 1, 2017 #4
    Sorry about that I should have been clearer. I meant x = ∫vdt. The formula holds even with a non constant velocity.

    Not quite. The situation you described requires that you break the movement into two parts: one where the object moves with velocity v, and the other when the object moves with velocity -v. In that case you have two displacements: from 0 to vt, which is positive just like v, and then from vt to vt1, which is negative just like -v. In both cases, you calculate the displacement by taking the final position minus the initial position. Your initial position differs in the calculations. The net effect is that your position becomes v(t - t1), but this is not your displacement This is the sum of your displacements. You have to account for the change in direction of the velocity. That's why you need to divide motion in two parts. In this situation your position is a non-linear function of time and you must account for that by dividing the displacement into different parts for each change in direction. If you did not, you would end up measuring the wrong amount of time to travel different displacements.
     
  6. Feb 1, 2017 #5

    ehild

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    You are right, the velocity and the displacement need not have the same direction. The velocity is the time derivative of displacement.
    As an example, the displacement of a stone thrown up is Δy(t)=10t - 5 t2, the velocity is v=10 - 10t. At the instant t = 0.2 s, both the displacement and the velocity are positive. At the instant t=1, the displacement is positive and the velocity is zero. At t=1.5, the displacement is positive and v is negative.
    The average velocity in a time interval Δt is defined as the displacement in that interval divided by the length of the interval vavr=Δx/Δt. The direction of the average velocity is the same as that of the displacement in that time interval.
     
  7. Feb 1, 2017 #6
    Sorry about the confusion. I was thinking of the curve you get when you draw the position. In that case, you draw a tangent to the curve and the gradient of that tangent is your velocity. Then for a small displacement the velocity would have to be in the same direction. Again I apologise for the confusion.
     
  8. Feb 1, 2017 #7
    In your return journey the distance between you and the starting point decreasing hence it should have negative sign
     
  9. Feb 1, 2017 #8

    CWatters

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    .

    No that's incorrect. The "distance between you and the starting point" is still positive. See post #5. If you define ground level as the origin and up as positive then the "distance between you and the starting point" is always positive even on the way back down. It would only be negative below ground level.
     
  10. Feb 1, 2017 #9
    Well this thread is becoming very contradictory, and it's my fault; I'll try and clear up things.

    There is the position. Position is the coordinates of your point relative to (0,0,0) the origin. In a sense, it is the displacement from the origin. I'll designate it by p.
    Then there is displacement. Displacement is defined as the change in position. It is the current position minus the initial position. I'll designate it as s. It is:
    s = p - po
    where po is the initial position.

    Finally there is the infinitesimal change in position. This is what is used to define the instantaneous velocity. It is the position a time dt from now minus the current position. Graphically, it is the secant as the two points get closer and closer. The direction of this quantity is always that of the velocity. Dividing this quantity by dt will give you the velocity. It is represented by ds
     
  11. Feb 1, 2017 #10
    Consider an upward and downward travel, in the upward direction displacement and velocity are both positive
    In downward direction at time t you are in 'y' distance at t+dt time you are at 'y+dy' distance, the decrease in distance is equal to y-(y+dy) =-dy
    Velocity = -dy/dt
    Therefore the displacement and velocity are both in negative direction, the text book is correct
     
  12. Feb 1, 2017 #11
    The quantity y - (y + dy) is not the displacement. The displacement is y - yo where yo is the initial position. If we set the initial position as the origin, then the displacement is y. In that case the displacement is positive, and the velocity is negative. What you are describing is what I called the infinitesimal change in position. Although seen the number of disagreements this has caused, I'm beginning to wonder if there is not two different definitions of displacement …
     
  13. Feb 1, 2017 #12

    CWatters

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  14. Feb 2, 2017 #13
    Thank you all for the replies. Seems to have generated discussion. I think we can all agree though that my textbook could have been a bit more clear.

    FYI, this is an Australian Year 12 (VCE) textbook.
     
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