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Vectors with velocity and acceleration

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    An object moves with constant acceleration a = 1.9 m/s2 in the due East direction. At time t = 0, the velocity of the object is v0 = 2.2 m/s in the due North direction. What is the magnitude of the velocity at t = 2 s

    a]v=2.9 m/2
    b]v = 3.6 m/s
    c]v = 4.1 m/s
    d]v = 4.4 m/s
    e]v = 6.0 m/s

    3. The attempt at a solution

    I am really lost on this one, my manual for this class doesn't seem to tell me anything about this

    Would I have to find the components of the velocity or acceleration and then use vf=Vo+at?

    I tried calculating both the x and y values of the velocity but that didn't help me at all when i plugged them into the equation
    Last edited: Sep 16, 2010
  2. jcsd
  3. Sep 16, 2010 #2
    Hello! I'd like to try and help. I hope it is right!

    We have two directions, North and East. Thus East = x axis, North = y axis. I'm sure you had that, but just saying.

    We know that the resultant a is in the east direction and is 1.9 m/s^2. Thus, the y component is zero. So a = ax.

    We also know that Vi is 2.2 m/s due north, and thus Vix = 0 m/s, so Viy = Vi.

    We can get the velocity resultant by integrating a = 1.9 m/s^2. What would the constant of equation represent? We can then plug in t = 2s and hopefully get the correct answer.
  4. Sep 16, 2010 #3
    Thanks for your help, I got as far as the fourth line, however this class is take with Calc I, so the last line and the whole topic of integration makes no sense to me, or is it just that I'm unfamiliar with the terms you are using?
  5. Sep 16, 2010 #4
    Ah. That is fine! You haven't covered integration in Calc 1 yet. Which is fine. I'm guessing this is still calc based physics and therefore it's ok to use calculus in your answers.

    You have learned about derivatives a bit? Integrals are often referred to as an "anti-derivative". So a = a constant. What kind of derivative would give us that answer?

    Perhaps there is a way to do it algebraically, I just looked at it this way first. . .

    Haha! there is a way to do it algebraically. You already have the equation!
    We have vi, a, and t. We can solve for Vf at t = 2 sec
  6. Sep 16, 2010 #5
    So to obtain the final answer I would just use vf=vo+at?

  7. Sep 16, 2010 #6
    It seems so. . . we are looking for the magnitude of the velocity resultant, so we aren't concerned with the new direction.

    I'm getting all paranoid because it seems simple lol. Good luck!
  8. Sep 16, 2010 #7
    Thanks for your help. That's what I originally had but thought it was too simple as well!
  9. Sep 16, 2010 #8
    You are mixing components here. The acceleration in along the x axis (east) and the vo=2.2 m/s is along the y axis (north).
    You need to apply the formula for each axis, find the two components of the velocity at time t and then find the magnitude.
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