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find the magnitude of average accleration of the tip of second hand of length 10 cm during 10 seconds....

Physics
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there are some assumptions which you should make is that, the seconds have move with constant angular velocity 'w' time period is 2pi/w=60. w=2pi/60 10 secs means it travels 1/6th part of the clock.|dw:1382102831264:dw| it goes from A to B. change in velocity and solve like the previous case how we did it. and then that divide by the total time taken=10secs. do this and you'll get the ans.
Radial acceleration is \(R\omega^2\) assuming constant angular velocity. The angular velocity is \(\dfrac{2\pi}{60} \rm rad s^{-1}\)
@AllTehMaffs pls..hlp me in this question

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Other answers:

ahhww.. i dunno how to even begin.. average acceleration shoudl be zero cause before 10 seconds and after 10 seconds, the second hand comes to stop :-/
oh wait.. are you talking about centripetal acceleration!?
then what parth says is right
if it's right then how we r going to calculate acceleration
i m convinced with this that linear velocity is cuming as pi/300 m/sec
then u know what is the relation between linear vel and centripetal acceleration
it is v^2/r
there you ahve it.. plug it in!
it's cuming as pi^2/90
no no sory i made a mistake lemme solve it
it's pi^2/9000
yea.. thats right..
u knw u can solve this simply.. 60 seconds --- 2 pi rads 1 second -- pi/30 rads so a = w^2 r = (pi^2/900)*0.1
bt the answer is given as pi/3000 m/s^2
here have they taken the value of one pi as 3 and then solve it
if dey hve done like this then it wud give the same answer to our question also am i ryt....?
The answer is π²/9000
ryt my answer is also that only
That's also the answer Mashy gave you.
the answer given is correct. it's pi/3000 m/s^2. at least this is what i'm getting.
how r u getting this answer cuz i m getting it as pi^2/9000
Hi @samigupta8, I am so sorry, but @Mashy and myself have led you to a wrong solution to your problem. Actually, @Abhishek619 gave you the right answer at the very beginning. As the problem asks for AVERAGE acceleration from position 1 to position 2, as acceleration is a vector, the right answer is: \(\vec a_{average} = \Large \frac {\vec v_2-\vec v_1}{t_2-t_1}\) with \(\Delta t = t_2-t_1\)=10 s and \(\vec v_1\) and \(\vec v_2\) as drawn in Abhishek619's sketch. |dw:1382428975483:dw| Hence \(\vec v_2-\vec v_1\) has the same magnitude v as v1 and v2 because all angles are π/3 the magnitude of the average acceleration is a = v/\(\Delta t\) Since v = 2πr/60s = 2π x 0.1 /60 = π / 300 m/s and \(\Delta t\) = 10 s then average acceleration is a = π / 3000 m/s²
drats.. thanks a lot for that update man!!

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