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idku

  • one year ago

I need help with basic question. (I need the reasoning not the answer)

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  1. idku
    • one year ago
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    At t=0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. You notice it moves 1 foot between t=0 seconds and t = 1 second. 1) How far does it move between t = 1 second and t = 2 seconds?

  2. UnkleRhaukus
    • one year ago
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    \(\theta\)|dw:1440728525844:dw|

  3. UnkleRhaukus
    • one year ago
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    |dw:1440728672933:dw|

  4. UnkleRhaukus
    • one year ago
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    |dw:1440728734123:dw|

  5. idku
    • one year ago
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    I haven't ever done anything like that. Couldn't we relate to the fact that acceleration is the second derivative of the displacement?

  6. idku
    • one year ago
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    MY teacher hasn't gone through anything of this sort in class. I am really sorry that you typed all of that.

  7. UnkleRhaukus
    • one year ago
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    \[W_y+N=0\]

  8. UnkleRhaukus
    • one year ago
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    \[\sum F = W+N\\ m\ddot x= mg \sin\theta \]

  9. idku
    • one year ago
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    d=\(\dfrac{1}{2}\)\(\times\)a\(\times\)t\(^2\) this is the relation to an acceleration in terms of a, am I correct? (where t is time)

  10. UnkleRhaukus
    • one year ago
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    yeah, and we have \[a = \ddot x = g\sin \theta\]

  11. idku
    • one year ago
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    I really have no clue about the x with two dots on top-:(

  12. UnkleRhaukus
    • one year ago
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    the dots are time derivatives \[\ddot x = \frac{\mathrm d^2x}{\mathrm dt^2}\]

  13. idku
    • one year ago
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    x=\(\dfrac{1}{2}\)at\(^2\) when x=1 and t=1 (as in our case) 1=\(\dfrac{1}{2}\)a(1)\(^2\) --> a=2 Am I going in the correct direction? I am just trying to apply that one fact about the acceleration that it is 2nd derivative of positon with respect to time.

  14. idku
    • one year ago
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    (x=1 and t=1, because it goes a distance of 1 foot from t=0 to t=1)

  15. idku
    • one year ago
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    Now I will find the distance the ball travelled from t=0 to t=2 (I know t=2 because I am looking for displacement after 2 second, and I know a=2 because that is my acceleration) x=\(\dfrac{1}{2}\)(2)(2)\(^2\) x=4

  16. idku
    • one year ago
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    So I found that from t=0 to t=1 it travelled 1 foot, and from t=0 to t=2 it travelled 4 feet. So, between t=1 and t=2 it travelled 4-1=3 feet.

  17. UnkleRhaukus
    • one year ago
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    We found the acceleration to be (a constant) \[a(t) = g\sin\theta\] Integrating this twice, with respect to time: \[x(t) = \iint a(t)\,\mathrm dt\,\mathrm dt\\ = g\sin theta\iint \,\mathrm dt\,\mathrm dt\\ = g\sin\theta t^2/2+c\]

  18. idku
    • one year ago
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    I am stupid, indeed. I like the more elementary way-:( I don't understand what you accomplished, and for you it is obvious as daylight.

  19. UnkleRhaukus
    • one year ago
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    setting the origin to be at x(t=0) = 0 implies \[ x(t) = g\sin\theta\cdot\frac {t^2}2\]

  20. UnkleRhaukus
    • one year ago
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    \[x(t=1)=g\sin\theta \cdot(1)^2/2=1[\text{ft}]\]

  21. UnkleRhaukus
    • one year ago
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    so \[g\sin\theta/2 =1\] \[g\sin\theta =2\] \[\therefore a= 2\]

  22. UnkleRhaukus
    • one year ago
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    \[\implies\qquad x(t) = t^2\]

  23. UnkleRhaukus
    • one year ago
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    yes your right, it it moves 3 [ft] from t=1 to t=2

  24. idku
    • one year ago
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    I, for my huge math inability, when I left a paper from calculus I my relatives were saying that I am a genius, even though it is fairly simple. Same thing is when you proposed your approach with is simple to you, but to me....

  25. idku
    • one year ago
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    Thanks for your input though.... some time later, I will probably recall you posted something like this and get back to it when my knowledge will be efficient for that.

  26. idku
    • one year ago
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    I am just poor at maths, sorry-:(

  27. UnkleRhaukus
    • one year ago
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    i was looking at the physics, (it wasn't really necessary, but you asked for reasoning )

  28. idku
    • one year ago
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    Well, what I said contained reasoning. Or didn't it? (My head is going crazy after a wild 16h day. Apologize again)

  29. UnkleRhaukus
    • one year ago
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    Are all the necessary understandings clear now?

  30. UnkleRhaukus
    • one year ago
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    You method was simple acceleration is constant, so the position changes as x(t) = at^2/2 you used the point (t=1[s], x=1[ft]) to find the value of a=2 and then found the value at t=2[s] with x(t) = at^2/2 =t^2 the distance traveled between t=1&2 was the difference x(t=1[s]) - x(t=1[s])

  31. UnkleRhaukus
    • one year ago
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    I can find the angle \(\theta\), from the diagram, where the slope is at an angle \(\theta\) \[a = g\sin theta\] gravity \(g\approx10\) and we found acceleration to be \(a=2\) so\[2=10\sin\theta\qquad\implies \theta = \arcsin(1/5) \approx 11.5°\]

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