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alphabeta

  • one year ago

How to find the instantaneous acceleration by looking at this graph (file attached) at time 0.50s?

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  1. alphabeta
    • one year ago
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  2. Abhisar
    • one year ago
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    |dw:1442065747164:dw| Hey alphabeta! Acceleration is the rate of change in velocity or in other words, \(\sf acceleration = \Large \frac{dv}{dt}\)= Slope of Velocity-time graph. So, let's suppose we want to calculate the acceleration of the body at t=1 sec, then slope will roughly be equal to 6 over 1 = \(\sf \Large \frac{6}{1}\)

  3. alphabeta
    • one year ago
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    Thanks @Abhisar ! That really helped a lot! :-)

  4. Abhisar
    • one year ago
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    You're welcome c:

  5. alphabeta
    • one year ago
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    At least I thought I understood? For some reason, I don't get the right answer! I'm doing it like this: (4.2ms^-1)/(0.50s) = 8.4 ms^-2, but the markscheme says 4.5ms^-2... But when thinking about it, isn't this the average acceleration rather than the instantaneous?

  6. alphabeta
    • one year ago
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    @Abhisar

  7. Abhisar
    • one year ago
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    What do you want to figure out, instantaneous acceleration or average acceleration?

  8. alphabeta
    • one year ago
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    Instantaneous (as written in my first post) :-)

  9. Abhisar
    • one year ago
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    Well, in that case it should be 8.4 m/s^2. That's what I believe but let's call @IrishBoy123 to counter check it.

  10. IrishBoy123
    • one year ago
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    |dw:1442072505379:dw|you need the slope of the tangent line at t = 0.5 i think @Abhisar 's line is parallel and gives you a very good approximation so your denominator should be 1 in your calculation but it's just an approximation, isn't it :p

  11. IrishBoy123
    • one year ago
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    that's for **instantaneous acceleration** at t = 0.5, to be clear. the slope at that instant

  12. Abhisar
    • one year ago
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    Yeah, basically it needs to be a tangent drawn through the curve as you have drawn.

  13. Abhisar
    • one year ago
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    Thanks c:

  14. Astrophysics
    • one year ago
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    |dw:1442092345797:dw| I am approximating as we don't have a computer to be exact, and as you were still confused maybe this will clear things up a bit @alphabeta this is for instantaneous acceleration, keeping it simple \[a = \frac{ \Delta v }{ \Delta t }\] notice we take the slope of the line here at that instant.

  15. alphabeta
    • one year ago
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    Ahh, now I see! Thanks soo much! :-D

  16. Astrophysics
    • one year ago
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    Yw :)

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