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ParthKohli

  • 2 years ago

Classical mechanics.

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  1. ParthKohli
    • 2 years ago
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    |dw:1370881868506:dw|At what time will the particles collide?

  2. ParthKohli
    • 2 years ago
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    @amistre64

  3. ParthKohli
    • 2 years ago
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    |dw:1370882089218:dw|

  4. ParthKohli
    • 2 years ago
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    I tried this:\[s_1 = 4t + \dfrac{1}{2}t^2\]\[s_2 = 2t + \dfrac{1}{2}t^2\]And \(s_1 + s_2 = 30\) so\[t^2 + 6t = 30 \Rightarrow t^2 + 6t - 30 = 0\]

  5. ParthKohli
    • 2 years ago
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    But am I right?

  6. amistre64
    • 2 years ago
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    i see that is what, acceleration up top and speed on the bottom?

  7. ParthKohli
    • 2 years ago
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    Yes.

  8. amistre64
    • 2 years ago
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    \[\frac12at^2+v_it+d_i=\frac12\alpha t^2+\beta_it+\delta d_i\] \[\frac12t^2+4t+0=-\frac12t^2+2t+30\]

  9. amistre64
    • 2 years ago
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    im wondering it that should be a -2t over on the right

  10. ParthKohli
    • 2 years ago
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    That is what I have also been confused about for a while, but I figured that it should be a \(+\) because we're dealing with distances and not displacements. If we were dealing with displacements, \(s_1 + s_2\) would not have been \(30\) either.

  11. amistre64
    • 2 years ago
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    since velocity is a vector, all assume -2t \[\frac12t^2+4t=-\frac12t^2-2t+30\] \[t^2+6t=30\]

  12. ParthKohli
    • 2 years ago
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    Well, so that leads to \(t^2 + 6t - 30\)... so I am right?

  13. ParthKohli
    • 2 years ago
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    Thanks, another one coming up: this time with circles!

  14. Vincent-Lyon.Fr
    • 2 years ago
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    Yes this is correct. Choose solution with t>0.

  15. ParthKohli
    • 2 years ago
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    Yes, already done. Thank you

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