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What's the car's acceleration at specific point?

  1. Feb 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots" of the car's position at equal 0.5 s time intervals. (Assume the positive direction is to the right. Indicate the direction with the sign of your answer.) t=1 s 0.4 m and then t=1.5 s 0.9m and then t=2 s 1.6 m.

    Using data from t = 1.0 s to t = 2.0 s, what is the car's acceleration at t = 1.5 s?

    2. Relevant equations
    I know average velocity equals Delta v/Delta t but I'm confused when it's about a specific point.

    3. The attempt at a solution
    I tried 1.4 (Vf) - 1(Vi)/(2-1) and got 0.4 m/s which was wrong
     
  2. jcsd
  3. Feb 9, 2015 #2

    lightgrav

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    yes, vf=1.4 (m/s) ... the units go after the numbers, because the numbers are multiplied by those units.
    but your vi is wrong ... so your acceleration is wrong (also you wrote wrong units for acceleration).
     
  4. Feb 9, 2015 #3
    My initial velocity is wrong? it's not 1? Wouldn't it be .5/.5=1? So then 1.4-1 within the acceleration equation. I guess I'm a bit confused what I'm doing wrong
     
  5. Feb 9, 2015 #4

    lightgrav

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    Sorry, you were using peculiar notation (and without proper units) so I mis-understood what you wrote.
    Do you think you got it marked wrong for the units being wrong?
     
  6. Feb 9, 2015 #5
    No, it's online so it automatically assigns the units (that's why I don't usually pay attention to them). Because if I did the formula correctly wouldn't it be 0.4? When I get it wrong it says "Recall that the definition of average acceleration is the change in velocity divided by the change in time. What is the initial velocity? What is the final velocity? Be careful about the change in time as well—if you measure velocities at "midpoints" within two separate intervals, what is the separation between those midpoints? m/s2."
     
  7. Feb 9, 2015 #6

    Nathanael

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    I get a different answer than you. If you assume acceleration is constant from t=1 second to t=2 seconds, then you can write two equations for the distance. (Call the velocity at time t=1 second "v" and call the acceleration "a" and write two equations then solve for v and a)

    EDIT:
    I like your way better, it's more straightforward. I didn't understand your technique until I read gneill's comment
     
    Last edited: Feb 9, 2015
  8. Feb 9, 2015 #7

    gneill

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    Staff: Mentor

    Shouldn't the time interval be 1/2 second?
     
  9. Feb 9, 2015 #8
    This is the figure with the problem. I don't think acceleration is constant in this problem? Do you think it's asking for the instantaneous acceleration, maybe?
     

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  10. Feb 9, 2015 #9

    Nathanael

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    Check for yourself if it's constant. It's easier with the additional information from the picture. Use the constant acceleration (from rest) equation [distance=at2/2] and see if it gives the same value of acceleration for each distance. If it does, then acceleration is constant. If not, then it's not.
     
  11. Feb 9, 2015 #10
    I got it! You guys were trying to tell me that the time intervals should have been 2-1.5 and not 2-1. I thought because it said "use the info between t=1 and t=2 to solve the problem" that I had to use t=1 and t=2. But I just did 1.4-1/2-.15 and got 0.8 and that was correct :).

    I think I got hung up on the fact that it said at t=1.5 where the earlier question within it said between 1 and 1.5

    Thanks for the help!
     
  12. Feb 9, 2015 #11

    Nathanael

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    Hmm... I understand it differently.

    Your equation was [itex]a=\frac{V_f-V_i}{Δt}=\frac{1.4-1}{Δt}[/itex] right?
    If I understand correctly, the "Vi=1" comes from [itex]\frac{0.9-0.4}{0.5}=[/itex]Average velocity from t=1 to t=1.5
    And the "Vf=1.4"comes from [itex]\frac{1.6-0.9}{0.5}=[/itex]Average velocity from t=1.5 to t=2

    Now the thing about average velocities with constant acceleration is that the average velocity is equal to the instantaneous velocity at the time half way between the initial and final times.
    So the "Vi=1" happens at t=1.25 seconds and the "Vf=1.4" happens at t=1.75 seconds (therefore Δt=0.5)

    Even though the answer comes out the same, my point is that the time interval for your equation is from 1.25 seconds to 1.75 seconds, not from 1.5 seconds to 2 seconds.
     
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