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alyssajobug

  • one year ago

Please help, I'm not sure how to even approach this question... A Porsche accelerates from a stop light at 5.0m/s^2 for 5s, and then decelerates at 5.0m/s^2 for 3s. How far has it traveled?

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  1. anonymous
    • one year ago
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    Alright no problem

  2. anonymous
    • one year ago
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    d=0.5at^2

  3. anonymous
    • one year ago
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    use the above equation to find the displacement

  4. alyssajobug
    • one year ago
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    So I would do d=0.5 x 5.0m/s^2? or somewhere else?

  5. anonymous
    • one year ago
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    First one is the positive vector and negative for the second.

  6. anonymous
    • one year ago
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    Alright let me walk you through

  7. anonymous
    • one year ago
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    For the first displacement you can calcaulate the magnitude of displacement using 0 d=0.5(5m/s)(5s)^2 Notice how the units of seconds cancel out and what's left is "m" which accounts for the displacement.

  8. anonymous
    • one year ago
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    You do that for the first half of the question and then a=0.5(5m/s^2)(3m)^2 Subtract the first with the second to get the overall displacement.

  9. anonymous
    • one year ago
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    Notice that displacement only concerns the overall travelled distance from the initial position to your final position.

  10. anonymous
    • one year ago
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    I hope this helps. Let me know where you lack comprehenion .

  11. alyssajobug
    • one year ago
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    So do you subtract the first from the second then?

  12. alyssajobug
    • one year ago
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    So when solving for this I take the 7.5 that I got from the second equation and subtract it from the 12.5 I got from the first equation to get 5m^2 right? or did I misunderstand? Robert136?

  13. anonymous
    • one year ago
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    What you got is an overall displacement.

  14. anonymous
    • one year ago
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    For the first accelerating part you can use that equation because it's starting from rest. I don't think you can use it for decelerating part because it has velocity.

  15. anonymous
    • one year ago
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    You can use v = v0 + at to find the velocity at the end of the acceleration. Then use d = 0.5at² + vt to find the displacement for that part

  16. anonymous
    • one year ago
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    @peachp Thanks for your correction I totally forgot to account for the preexisting velocity

  17. anonymous
    • one year ago
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    make sure to use negative acceleration too

  18. anonymous
    • one year ago
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    no problem I meant use d = 0.5at² + vt to find the displacement for the deceleration part. Where a will be -5. And then you'd add the 2 displacements to get the overall

  19. anonymous
    • one year ago
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    @peachpi Truely so. It would have been wiser first calculate the initial velocity and then account for the negative acceleration. However the counterintuitive question is whether to fully consider the initial velocity as being part of the whole but as far as the equation goes negative acceleration looks after it all

  20. anonymous
    • one year ago
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    Physics always goes against our intuition.

  21. anonymous
    • one year ago
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    d=(25m/s*5s)+(-5m/s^2)(5s)^2 Should get you the overall displacement. Initial velocity was calculated using at=v

  22. anonymous
    • one year ago
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    Acceleration is change of velocity per unit of time so don't worry too much about always having to express it with seconds

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