#### Answer

(a) $a = 0.057~m/s^2$
(b) The length of the chain that passes over the sprocket is 7.85 meters.

#### Work Step by Step

Initially, the crank arm rotates at 60 rpm which is 1 rev/s which is $2\pi~rad/s$. Therefore $\omega_0 = 2\pi~rad/s$.
After ten seconds, the crank arm rotates at 90 rpm which is 1.5 rev/s which is $3\pi~rad/s$. Therefore $\omega_f = 3\pi~rad/s$.
We can find the angular acceleration of the crank arm as;
$\alpha = \frac{\omega_f-\omega_0}{t}$
$\alpha = \frac{3\pi~rad/s-2\pi~rad/s}{10~s}$
$\alpha = 0.314~rad/s^2$
We then find the tangential acceleration of the pedal;
$a = \alpha~r$
$a = (0.314~rad/s^2)(0.18~m)$
$a = 0.057~m/s^2$
(b) We can find $\theta$ during the 10-second acceleration period. Note that the angular acceleration of the sprocket is the same as the angular acceleration of the crank arm.
$\theta = \omega_0~t+\frac{1}{2}\alpha~t^2$
$\theta = (2\pi~rad/s)(10~s)+\frac{1}{2}(0.314~rad/s^2)(10~s)^2$
$\theta = 78.5~rad$
We first find the distance $x$ that a point on the edge of the sprocket rotates. Note that this distance is equal to the length of the chain that passes over the sprocket.
$x = \theta ~r$
$x = (78.5~rad)(0.10~m)$
$x = 7.85~m$
The length of the chain that passes over the sprocket is 7.85 meters.