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gandalfwiz
Anybody here good at physics and curious to know how long it takes a speeding Moped to reach Ke$ha?
The moped accelerates uniformly over a time of 5.21 seconds for a distance of 110 meters. I have to a) find the average velocity, b) find the velocity at 5.21 seconds, and c) determine acceleration of the Moped.
I'm pretty sure that the average velocity is 21.11 m/s (distance/time), but I'm unsure how to solve the others. Help?
does the moped start from rest?
My formula chart sez \[\Delta d = v_i \Delta t + \frac{1}{2}a (\Delta t)^2\] and you know d, v_initial, and t, so it's easy to solve for "a" ...
average velocity= (u+v)\2=21.21 m\s v=u+at u+v=42.22 m\s
final velocity is pretty fast, can we hope that Ke$ha won't be able to get out of the way of the speeding moped?
42.22-u=u+at 2u=42.22-5.21 a
lol @ fixer... but your formula is way too intense for me :) I think that there's a way to figure it out without one other than the basics...
ravi: what do u and v stand for?
i think ravi is saying that final velocity is twice average velocity (since initial velocity is zero), and than acceleration is (final velocity) / (time).
a=(42.22-2u)\2 u=initial velocity v=final velocity a=acceleration
hold on... let me process that information...
ravi- how'd you get 21.21? My calculations give 21.11
sir, it was written by mistake
okay, so it is 21.11? and then from there you said that velocity=initial velocity + acceleration * time?
and after that was initial velocity + final velocity= 42.22 m/s. Where did 42.22 come from?
everyone whos good at physics are at physics group
no one in the physics group :P
distance traveled=(average velocity). time=110
im glad some one could help u
Ravi? Where's u go :(
u+v=42.22 v-u=5.21 a so adding these equation, we get 2v=42.22+5.21a
a=(v-u)\t if moped started from rest then a=v\t=v\5.11
now we have final equation for v v=( 42.22+5.11*(v\5.11) ) \ 2 solving this equation for v, we can find the value of final velocity .
here average velocity= (u+v)\2