a bicyclist starts from rest and accelerates at 2 rad/s2. In 10 seconds what is the angular velocity of the bike wheels and how many revolutions has the bike wheels completed?

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a bicyclist starts from rest and accelerates at 2 rad/s2. In 10 seconds what is the angular velocity of the bike wheels and how many revolutions has the bike wheels completed?

Physics
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I'm bad at physics :(

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Other answers:

me too :( thanks anyway my friend :D
i have an attempt at the solution :)
i'll post it?
Can I try?
:D okay
Just check....
one second I did a mistake......
use w^2=2*alpha*t where alpha=2, t=10
sqrt(40) is angular velocity Check it.... i am pretty sure
6.324 you mean? i got 20 rad/s
hmmm....
|dw:1439636369189:dw| post the actual question or a link ??!!
yes its 20.... we get s=100 and putting in v^2-u^2=2*aplha*s we get sqrt(400) which is 20
we get s=100 bu s=1/2*alpha*t^2
yes you are right @KarlaKalurky
haha @IrishBoy123 that's as nice joke :D
w=2*pi*f where f is the frequency
20=2*pi*f f=10/pi
what are the formulas to solve for no. of rev?
w=2*pi/T where w=20 we get T=pi/10 (for one cycle)
100 rad(1rev/(2pi)rad) = 15.9155 rev ?
hmmm.... let me think....
wait.... I got it... w=2*pi*f 20/2*pi=f 3.18 revolution in one second so 10 second, 10*3.18=31.8
This should be the answer @KarlaKalurky
thank you very much! :D
no problem
you need the angular analogues of the equations of motion \(v = u + at \\x = ut + \frac {1}{2} a t^2 \\ v^2 = u^2 + 2 ax\) which are \(\omega_2 = \omega_1 + \alpha t \\\theta = \omega_1t + \frac {1}{2} \alpha t^2 \\ \omega_2^2 = \omega_1^2 + 2 \alpha \theta\)

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