Question

I need help with question number four all parts ABC And d

Answer 1

a)

We know that the angular acceleration is related to the linear acceleration by:

`a=\alpha r`

where alpha is the angular acceleation and r is the radius, then, in this case we have:

`a=(1.7)(0.385)=0.6545`

Therefore the linear acceleration is 0.6545 meters per second per second.

b)

The linear velocity is related to the angular velocity by:

`v=\omega r`

where omega is the angular velocity, plugging the values we have and solving for the angular velocity we have:

`\begin{gathered} 10.4=(0.385)\omega \\ \omega=(10.4)/(0.385) \\ \omega=27.013 \end{gathered}`

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:

`a=(v_f-v_0)/(t)`

since he started at rest this means that the initial velocity is zero; plugging the values we know and solving for t we have:

`\begin{gathered} 0.6545=(10.4-0)/(t) \\ t=(10.4)/(0.6545) \\ t=15.89 \end{gathered}`

Hence it takes 15.89 seconds to reach this velocity. To determine how many radians the wheels turned we use the fact that:

`\theta=\theta_0+\omega_0t+(1)/(2)\alpha t^2`

then we have:

`\begin{gathered} \theta=0+0(15.89)+(1)/(2)(1.7)(15.89)^2 \\ \theta=214.618 \end{gathered}`

Therefore the wheels turned 214.618 radians.

d)

To determine how far the bycicle traveled in this time we use:

`x=x_0+(1)/(2)v_0t+(1)/(2)at^2`

then we have:

`\begin{gathered} x=0+0(15.89)+(1)/(2)(0.6545)(15.89)^2 \\ x=82.628 \end{gathered}`

Therefore the bicycle traveled 82.628 meters.