a)
We know that the angular acceleration is related to the linear acceleration by:
[tex]a=\alpha r[/tex]
where alpha is the angular acceleation and r is the radius, then, in this case we have:
[tex]a=(1.7)(0.385)=0.6545[/tex]
Therefore the linear acceleration is 0.6545 meters per second per second.
b)
The linear velocity is related to the angular velocity by:
[tex]v=\omega r[/tex]
where omega is the angular velocity, plugging the values we have and solving for the angular velocity we have:
[tex]\begin{gathered} 10.4=(0.385)\omega \\ \omega=\frac{10.4}{0.385} \\ \omega=27.013 \end{gathered}[/tex]
Therefore the angular velocity is 27.013 radians per second.
c)
To determine the time it takes for the cyclist to reach that velocity we use the equation:
[tex]a=\frac{v_f-v_0}{t}[/tex]
since he started at rest this means that the initial velocity is zero; plugging the values we know and solving for t we have:
[tex]\begin{gathered} 0.6545=\frac{10.4-0}{t} \\ t=\frac{10.4}{0.6545} \\ t=15.89 \end{gathered}[/tex]
Hence it takes 15.89 seconds to reach this velocity. To determine how many radians the wheels turned we use the fact that:
[tex]\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2[/tex]
then we have:
[tex]\begin{gathered} \theta=0+0(15.89)+\frac{1}{2}(1.7)(15.89)^2 \\ \theta=214.618 \end{gathered}[/tex]
Therefore the wheels turned 214.618 radians.
d)
To determine how far the bycicle traveled in this time we use:
[tex]x=x_0+\frac{1}{2}v_0t+\frac{1}{2}at^2[/tex]
then we have:
[tex]\begin{gathered} x=0+0(15.89)+\frac{1}{2}(0.6545)(15.89)^2 \\ x=82.628 \end{gathered}[/tex]
Therefore the bicycle traveled 82.628 meters.
