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ParthKohli
 3 years ago
Classical mechanics.
ParthKohli
 3 years ago
Classical mechanics.

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

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

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes this is correct. Choose solution with t>0.

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