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At what time is the velocity perpendicular to the acceleration

  1. Dec 9, 2016 #1
    1. The problem statement, all variables and given/known data:
    A mass is moving with constant acceleration in the x-y plane. At time t = 0s the mass’s position is 2mˆı+3mjˆ, the mass’s velocity is 3 m/s ˆı −4 m / s jˆ and the mass is subject to an acceleration of 1 m/ s ^2 ˆı + 1 m/ s^ 2 i jˆ.

    At what time is the velocity perpendicular to the acceleration?

    2. Relevant equations:

    v = v0 + α t
    x = x0 + v0t + 0.5at^2
    v^2= v0^2 +2a(x-x0)

    3. The attempt at a solution:

    I don't really know how to approach this question. I tried to take the magnitude of velocity and acceleration and find the radius using a= v^2/R, but I don't think that is the way to solve it.

    I know that the answer is t=0.5 sec.

    Many thanks for your help

     
  2. jcsd
  3. Dec 9, 2016 #2
    In terms of the time t, what is the x component of velocity at time t? In terms of the time t, what is the y component of velocity at time t? In terms of the time t, what is the velocity vector at time t? If two vectors are perpendicular, what can we say about their dot product?
     
  4. Dec 9, 2016 #3

    haruspex

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    In vectors, what is the mass's velocity at time t?
     
  5. Dec 9, 2016 #4
    V(t)x= 3+t
    V(t)y= -4+t
    V = √(3+t)^2+(t-4)^2 , is that correct ?
    A = √2
    dot product: V ⋅ A = 0 because the angle is 90 deg.
    But When Im trying to solve it I get an unsolvable quadratic equation...
     
  6. Dec 9, 2016 #5

    nrqed

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    You need to work with vectors. What is the vector ##\vec{v(t)}##? What is the acceleration vector ##\vec{a}##?
     
  7. Dec 9, 2016 #6
    I think I solved it, but I am not sure I did it correctly:

    Vx(t) = 3+t
    a x = 1

    Vy(t) = -4+t
    ay=1

    cross product:
    (3+t)⋅1 + (t-4)⋅1=0
    t=0.5 sec
     
  8. Dec 9, 2016 #7

    haruspex

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    Right, but you mean dot product.
     
  9. Dec 9, 2016 #8

    nrqed

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    Yes, that's correct (note: you mean the *dot* product, not the cross product).
     
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