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ParthKohliBest ResponseYou've already chosen the best response.0
dw:1370881868506:dwAt what time will the particles collide?
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
dw:1370882089218:dw
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
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\[t^2 + 6t = 30 \Rightarrow t^2 + 6t  30 = 0\]
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
i see that is what, acceleration up top and speed on the bottom?
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
\[\frac12at^2+v_it+d_i=\frac12\alpha t^2+\beta_it+\delta d_i\] \[\frac12t^2+4t+0=\frac12t^2+2t+30\]
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
im wondering it that should be a 2t over on the right
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
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.
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
since velocity is a vector, all assume 2t \[\frac12t^2+4t=\frac12t^22t+30\] \[t^2+6t=30\]
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Well, so that leads to \(t^2 + 6t  30\)... so I am right?
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Thanks, another one coming up: this time with circles!
 10 months ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
Yes this is correct. Choose solution with t>0.
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Yes, already done. Thank you
 10 months ago
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