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An object moves with constant speed

  1. Feb 15, 2015 #1
    1. The problem statement, all variables and given/known data
    What relationship exists between the velocity and acceleration vectors? Prove your claim

    2. Relevant equations
    Since speed = |v| = constant, acceleration_tangential = d|v|/dt = 0. My question is on the "prove your claim part" Can I see simply just say let v(t) = a*i + b*t j (i = unit vector i, j = unit vector j). And since there is no t next to a or b, v(t) doesn't change and it's deritvative is zero ? Or do I have to show using dot product/cross product

    3. The attempt at a solution
     
  2. jcsd
  3. Feb 15, 2015 #2
    I think that's a bit too loose handwaving there. Try considering
    [tex]\frac{\mathrm{d}}{\mathrm{d}t} \left(\mathbf{v}^{2}\right) = \frac{\mathrm{d}}{\mathrm{d}t} \left(\mathbf{v}\cdot\mathbf{v}\right)[/tex]
    and see where that leads you to.
     
  4. Feb 15, 2015 #3
    Hi thanks for the quick response. I 'm not sure where to go with that. I know theres a theorem that says the dot product of a vector with itself is the magnitutde of the vector squared, but im not sure where I can go with what you said..

    But on the other hand, can I also use this formula , a = v' T + kv^2 N where v = speed , k= curvature, T = unit tangent ,, N= unit normal ,and since speed is constant v' is zero so a_tangential is 0 ?

    thanks
     
  5. Feb 15, 2015 #4

    PeroK

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    Why not differentiate the right-hand side? Assuming you know how to differentiate a dot product?
     
  6. Feb 15, 2015 #5
    I don't think that's a general formula. Actually, come to think of it, you have not formulated your claim clearly in the first place. The question asked you for the relationship between the acceleration and velocity vectors. All you have said is that [itex]\frac{d}{dt}|\mathbf{v}| = 0[/itex], which is just restating that the speed is constant.

    Yes. So [itex]\mathbf{v} \cdot \mathbf{v} = \mathbf{v}^{2}[/itex] is simply the speed squared, which is constant with time here. You will also need to recall that
    [itex]\mathbf{a} = \frac{d}{dt}\mathbf{v}[/itex]
     
  7. Feb 15, 2015 #6
    So d/dt (v • v) = v'•v + v•v' = 2 v•v' but I'm not sure where to go with this, is it that since the two vectors are orthogonal their dot product is thus zero and since this dot product = d/dt |v|^2 then that quantity is also zero , so acceleration is zero?


     
  8. Feb 15, 2015 #7

    PeroK

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    Go back a step. You have:

    ## \frac{d}{dt}(v^2) = 2 \mathbf{v.a} ##

    What do you know about the left hand side?
     
  9. Feb 15, 2015 #8
    That it is zero since |v| doesn't change. So a•v= 0 meaning either the vectors are orthogonal or one of the vectors is the zero vector ?
     
  10. Feb 15, 2015 #9

    PeroK

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    QED!
     
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