well i was tryign to make sense of it in the fact that
V= dy/dt
Vdt = dy, and dy = yf-yi or change in y
then ∫V dt = yf - yi is where i get stuck because how you can just integrate one side and how do you have the wherewithall to do this.
the kinematics equation i used upstairs; yf = yi +viy *t +1/2 ay t^2
also how does ∫(dy/dt)dt = yf - yi
could i just shortcut and say ∫(dy/dt)dt = yf?
I think I'm utterly lost here...isn't dy/dt already vf-vi when change in time approaches 0
Yeah you're right, I guess here I just didn't understand how to differentiate (figuratively) the different of when I have constant a or constant v versus not, the thing is once I have the velocity that I got from integrating my acceleration, why can't i just plug that into my kinematics...
AHHH you're right, and a genius, and a scholar.
Acceleration is not constant, so we can't use those formulas for constant acceleration.
whenever they give you an a with respect to time, this is a variability acceleration, so in order to achieve V we must integrate a.
Thanks.
But now if I'm...
Simple velocity vector problem; IS MY SOLUTION MANUAL WRONG OR AM I?!
Homework Statement
A particle starts from the origin with velocity 5i at t=0, and moves in the xy plane with a varying acceleration given by a= 6 Sqrt(t) j, where t is in s.
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Homework Statement
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Dude, listen I did all that. they are incredibly easy to break down into components BECAUSE THEY ARE COMPONENT VECTORS, it's -3000j and -3000i the third vector can be broken down using sin and cosine. MY FIRST POST talked about doing it graphically and that the problem is i didn't know if I...
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