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Velocity & Acceleration Question

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Dave walks to the left and then turns around and walks to the right without pausing. At the exact moment that he turns around is
    a. velocity & acceleration zero
    b. velocity zero & acceleration non -zero
    c. velocity non-zero, acceleration zero
    d. velocity and acceleration non-zero
    e. all of the above depending on the situation


    2. Relevant equations



    3. The attempt at a solution

    I think his velocity is non-zero because he's still walking while his acceleration is zero because his velocity is still constant?

    Also, if a particle has a constant position it has a non-zero velocity? (because it has to be changing) I believe you have a non-zero velocity when the acceleration slope is zero and when an object falls near the surface of the Earth.
     
    Last edited by a moderator: Apr 10, 2013
  2. jcsd
  3. Apr 9, 2013 #2

    Simon Bridge

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    Take this back to the definition of velocity and acceleration ...

    displacement is a change in position;
    velocity is rate of change of displacement;
    acceleration is rate of change of velocity;
    all three are vectors - which affects what counts as a "change".
    ... you need to think what is involved in instantaneous changes.

    So - at the exact instant Dave turns around, is his displacement changing? Is his velocity changing?
    Put it another way - is there any way to change direction without accelerating?
     
  4. Apr 9, 2013 #3
    Ah is displacement changing because if we look at it on the x-axis he would have been going in the negative direction to the positive direction. I think. His velocity would be zero in the left direction when he changes displacement. So he has to accelerate since he is changing velocity. Therefore, velocity and acceleration are both non-zero?
     
  5. Apr 9, 2013 #4

    Simon Bridge

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    Hmmm... lets make it concrete.
    We first see Dave travelling right to left, to the right of the origin.

    So his initial displacement will be +x (from the origin), his speed will be -v (changing displacement in the -x direction), and lets give him a constant speed for this part of the journey to keep things simple, so the acceleration is zero.

    He reaches the origin at some time T - where he turns around and heads back the other way at velocity +v
    At time T, before he turns around,
    1. what is his displacement from the origin?
    2. what is his velocity?
    At time T, after he turns around,
    3. what is his displacement from the origin?
    4. what is his velocity?

    5. what is the change in displacement?
    6. what is the change in velocity?
     
  6. Apr 10, 2013 #5

    Why is his speed -v if he is going +x from the origin?

    1. +x
    2. -v
    3. -x
    4. +v
    5. -x -x = -2x
    6. final velocity - initial velocity = 2v
     
  7. Apr 10, 2013 #6

    Simon Bridge

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    OK - you can handle velocities just fine.
    Displacements need a bit of work.

    1. This tells me that when Dave is at the origin, his displacement from the origin is some distance x to the right of the origin?
    3. Now you are saying that when he is at the origin, his displacement from the origin is some distance x to the left of the origin?

    I think you are confusing the direction of a displacement with the direction Dave is facing.
    Displacement is always from some position. When the displacement is from the origin, then it is the same as the position vector... so it may be easier for you to think of it in terms of position vectors instead.

    Considering your answer to #6 - what can you say about the acceleration?
     
  8. Apr 22, 2013 #7
    Thank You for your help!! I got it! :]
     
  9. Apr 22, 2013 #8

    Simon Bridge

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    Well done - you'll find a lot of science is just asking the right questions.
     
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