ParthKohli
  • ParthKohli
Classical mechanics.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ParthKohli
  • ParthKohli
|dw:1370881868506:dw|At what time will the particles collide?
ParthKohli
  • ParthKohli
@amistre64
ParthKohli
  • ParthKohli
|dw:1370882089218:dw|

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ParthKohli
  • ParthKohli
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\[t^2 + 6t = 30 \Rightarrow t^2 + 6t - 30 = 0\]
ParthKohli
  • ParthKohli
But am I right?
amistre64
  • amistre64
i see that is what, acceleration up top and speed on the bottom?
ParthKohli
  • ParthKohli
Yes.
amistre64
  • amistre64
\[\frac12at^2+v_it+d_i=\frac12\alpha t^2+\beta_it+\delta d_i\] \[\frac12t^2+4t+0=-\frac12t^2+2t+30\]
amistre64
  • amistre64
im wondering it that should be a -2t over on the right
ParthKohli
  • ParthKohli
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.
amistre64
  • amistre64
since velocity is a vector, all assume -2t \[\frac12t^2+4t=-\frac12t^2-2t+30\] \[t^2+6t=30\]
ParthKohli
  • ParthKohli
Well, so that leads to \(t^2 + 6t - 30\)... so I am right?
ParthKohli
  • ParthKohli
Thanks, another one coming up: this time with circles!
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Yes this is correct. Choose solution with t>0.
ParthKohli
  • ParthKohli
Yes, already done. Thank you

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