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ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
dw:1370882795176:dwFind the time when the particles collide.
 one year ago

kutabs Group TitleBest ResponseYou've already chosen the best response.0
What is the thing written on the left side of the diagram?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
@kutabs 2 rad/s So this time I tried to do with the fact that \(\theta_1 + \theta_2 = \dfrac{3\pi}{2}\) and using the equations of motion.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
That isn't a problem at all. The problem is that there are more questions. I am not able to figure out \(a_R\) of each particle at the time of collision.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
is teh acceleration given for the 2rad/sec speed?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Ah, no worries, I get it. It's easy to find \(v\) for each particle after I get the \(t\). Thank you for the time guys.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
this is then the same as the last question really\[\frac12a_1t^2v_1t+d_1=\frac12a_2t^2+v_2t+d_2\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
\[\theta_1 = 2t \]\[\theta_2 = t + t^2\]Thank you, I can manage after that.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I was just having this little confusion that I cleared myself. :
 one year ago

kutabs Group TitleBest ResponseYou've already chosen the best response.0
Let them meet at the angle theta as shown dw:1370883098101:dw For 1: Covering pi/2+(pi/2theta)= omega*t For 2: Covering pi/2+theta= omega*t+1/2*alpha*(t^2)
 one year ago

kutabs Group TitleBest ResponseYou've already chosen the best response.0
Now from these two equations 1&2there are 2 variables theta and time (t). Solve them, and you'll get the result.
 one year ago
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