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Answer 1 · 2017-04-17T14:24:20

The speed of the particle is $dot s(t) = omega r = sqrt2 omega(t)$

For angular speed $omega$ , we note that:

$dot omega(t) = alpha(t) = pi/4$

$implies omega(t) = pi/4t + C$ and $omega(0) = 0 implies C = 0$

We also note that:

$dot theta(t) = omega(t) = pi/4 t$

$implies theta(t) = pi/8t^2 + C$ and $theta(0) = 0 implies C = 0$

From that, for a quarter turn in time $t = tau$ , we have:

$pi/2 = pi/8 tau^2 implies tau = 2$

Returning to speed $dot s$ , we now have: $dot s(t) = sqrt2 pi/4t$

The average speed $dot s_(ave)$ over the time period $tau$ is:

$dot s_(ave) = (int_0^tau dot s dt)/(tau)$

The numerator is just the distance travelled, ie a quarter turn along the circumference:

$= (pi/2 sqrt2 )/(2) = ( pi)/(2sqrt2)$