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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0dw:1370881868506:dwAt what time will the particles collide?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0dw:1370882089218:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I 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\[t^2 + 6t = 30 \Rightarrow t^2 + 6t  30 = 0\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i see that is what, acceleration up top and speed on the bottom?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac12at^2+v_it+d_i=\frac12\alpha t^2+\beta_it+\delta d_i\] \[\frac12t^2+4t+0=\frac12t^2+2t+30\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1im wondering it that should be a 2t over on the right

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0That 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
 one year ago
Best ResponseYou've already chosen the best response.1since velocity is a vector, all assume 2t \[\frac12t^2+4t=\frac12t^22t+30\] \[t^2+6t=30\]

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Thanks, another one coming up: this time with circles!

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.0Yes this is correct. Choose solution with t>0.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Yes, already done. Thank you
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