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Finding displacement and time from velocity and acceleration

  1. Oct 1, 2015 #1
    1. The problem statement, all variables and given/known data
    He decelerates uniformly at a rate of -5.00m/s^2 over from a velocity of 7.5m/s to rest. Find the time and displacement.

    2. Relevant equations
    Vf^2=Vo^2 + 2a(displacement)

    3. The attempt at a solution
    0=56.25+2(-5)x is this the right way to find displacement? Solving for the x?

    For time would it be 0 = 7.5 + (-5)(t) which is 1.5? That doesn't really seem correct I'm confused.
     
  2. jcsd
  3. Oct 1, 2015 #2

    Orodruin

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    Why do you think it is not correct?
     
  4. Oct 1, 2015 #3
    The time is done correctly. The displacement is not. That's because you chose the wrong equation. What is the equation that relates the displacement to the initial velocity, the acceleration, and the time?

    Chet
     
  5. Oct 1, 2015 #4
    Ok so if I did 1/2 (Vf + Vo)t would that be the correct one?
     
  6. Oct 1, 2015 #5
    I don't really know it just seemed so short
     
  7. Oct 1, 2015 #6
    That would be OK to use. I was thinking more of ##V_0t+\frac{1}{2}at^2##, which would give the same answer.
     
  8. Oct 1, 2015 #7
    Using this formula:
    [tex] 2a(\Delta x) = V_x^2 - V_0^2[/tex]
    Divide ## 2a ## to the other side
    [tex] x - 0 = \frac{(0^2 - (7.5m/s)^2)}{(2(-5m/s^2))} [/tex]
    [tex]x = \frac {-56.25m^2/s^2}{-10m/s^2}[/tex]
    Units:
    [tex]m = \frac {m^2}{s^2} * \frac{s^2}{m}[/tex]
    Left with meters:
    [tex]x = 5.625m[/tex]

    As far as with your time, does it make sense that you're starting out at 7.5 m/s and when one decelerates at ##-5.0m/s^2##, it happens decently fast?
    Using: ##V_x = a_xt + v_{0x}##
    @ 0.5s, ##V_x## = 5m/s
    ##V_x = (-5m/s^2)*0.5s + 7.5m/s##​
    @ 1s = 2.5 m/s
    ##V_x = (-5m/s^2)*0.5s + 7.5m/s##​
    @ 1.5s = 0
    ##V_x = (-5m/s^2)*0.5s + 7.5m/s##​
     
    Last edited: Oct 1, 2015
  9. Oct 3, 2015 #8

    gneill

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    @IntelvsAMD : Be careful about providing complete solutions to other's problems: it's against Physics Forum rules. You may give hints, suggestions, point out errors and so on, but the questioner must do their own homework and hopefully in the process learn how to do it themselves in future.
     
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