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anonymous

  • one year ago

Man has a varying displacements ranging from 1.0m,2.0m and 3.0m from his starting position and he had initial velocity of -2.0m/s and acceleration of 2.0m/s^2. Calculate the time elapsed at times when he is exactly at the 1.0m, 2.0m, 3.0m displacement from his starting point. Formula required. I am actually stuck with the formula that may or may not require quadratic solution to isolate (t). Anyone?

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  1. anonymous
    • one year ago
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    @Miracrown

  2. anonymous
    • one year ago
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    Excuse me the acceleration is 0.5m/s^2 actually

  3. Miracrown
    • one year ago
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    This is a physics Q, right?

  4. anonymous
    • one year ago
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    spot on sir

  5. anonymous
    • one year ago
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    can also be mathematically computed as well.

  6. anonymous
    • one year ago
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    It's hard to tell if a question should belong to physics or maths

  7. Miracrown
    • one year ago
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    Aww just how much I love being called ''Sir'' :D

  8. Miracrown
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Robert136 It's hard to tell if a question should belong to physics or maths \(\color{blue}{\text{End of Quote}}\) Nah mate, so easy to tell! But in some cases it may not be so easy. Your not wrong.

  9. Miracrown
    • one year ago
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    I have a question. Does the question mean displacements or positions of 1, 2, and 3 m?

  10. Miracrown
    • one year ago
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    Oh, it doesn't give an initial position, so it is displacement. So, can you write or type the quadratic equation you've found?

  11. anonymous
    • one year ago
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    Yeah the initial position is 0.00m right on

  12. anonymous
    • one year ago
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    It's supposed to go negative position for a while whilst the initial velocity holds negative but then due to acceleration it's turned to positive and eventually you know what's happening

  13. anonymous
    • one year ago
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    d=v(initial)*t+1/2(a)t^2 Where d=displacement or distance v=velocity t=time taken

  14. anonymous
    • one year ago
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    It's quid spot on how this equation is supposed to work to compute the distance or displacement.

  15. anonymous
    • one year ago
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    However, it's unclear as to whether d stands for displacement or distance in this given situation

  16. Miracrown
    • one year ago
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    This equation is for displacement. So, here's what's awesome about it. You don't have to worry about the fact that the man has an initial negative velocity and will have negative displacements. All you have to do is use d = 1.0, 2.0, and 3.0 m.

  17. anonymous
    • one year ago
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    So I plug in any of the given values being investigated and leave out the t and t squared, which is bit of a problem because then quadratic formula needs a peek

  18. Miracrown
    • one year ago
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    It's great that you recognize what the guy is actually doing, though! :D Right, you don't substitute anything for t because that's what you're solving for.

  19. anonymous
    • one year ago
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    Use the equation \[\vec s=\vec ut+\frac{1}{2}\vec a t^2\] As this looks like a question in 1D, you can simply replace vectors with scalars accompanied by appropriate plus or minus sign

  20. anonymous
    • one year ago
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    the vector s is displacement

  21. Miracrown
    • one year ago
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    d = v0 * t + (1/2)a*t^2 1.0 = -2t + (1/2)(0.5)t^2 0 = -1 - 2t + 0.25t^2 And once you have this, you can solve using the quadratic formula or by graphing.

  22. anonymous
    • one year ago
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    Here's the deal, as you are dealing in 1 dimension, you can only move forward or backward, like along a line. So you can assign either the forward or the backward direction a positive sign and it's opposite direction as negative so if a quantity is given as positive and another quantity is given as negative, what they are basically saying is they are in the opposite direction

  23. anonymous
    • one year ago
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    That's indicated by positive and negative sign here.

  24. Miracrown
    • one year ago
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    To be clear, when I say "graph it", I mean you'll input y = -1 - 2t + 0.25t^2 and the find its zeroes and your calculator will use x instead of t

  25. anonymous
    • one year ago
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    Right. I need to find the roots which would have two obviously but I need to pick the positive one because negative one will be unrealistic to even assume it exists.

  26. anonymous
    • one year ago
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    That's why I love physics

  27. anonymous
    • one year ago
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    relatively less math to deal with but with more intuitive reasoning

  28. anonymous
    • one year ago
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    XD

  29. Miracrown
    • one year ago
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    Physics is the mother of science

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