A wheel of radius a rolls along a straight line, turning

A wheel of radius $a$ rolls along a straight line, turning through $\varphi = t+\dfrac{t^{2}}{2}$ radians in $t$ seconds. Find the velocity and acceleration of its centre.

Step 1: The rolling condition

Rolling without slipping means the arc that has touched the ground equals the distance advanced, so the centre moves through

\[ s(t) = a\varphi(t) = a\left(t+\frac{t^{2}}{2}\right). \]

Step 2: Velocity

\[ v = \frac{ds}{dt} = a\frac{d\varphi}{dt} = a\bigl(1+t\bigr) = a\omega, \qquad \omega = 1+t. \]

Step 3: Acceleration

\[ w = \frac{dv}{dt} = a\frac{d^{2}\varphi}{dt^{2}} = a\cdot1 = a. \]

The angular acceleration is $\varepsilon = 1\ \text{rad/s}^{2}$, constant, so the centre's acceleration $a\varepsilon = a$ is constant too.

Step 4: Verification

\[t = 0:\ \varphi = 0,\ v = a,\ \text{the wheel is already turning at } 1\ \text{rad/s} \;\;\checkmark\]

\[\text{Distance check: } s(2) = a\bigl(2+2\bigr) = 4a; \text{ and } \int_{0}^{2}a(1+t)\,dt = a\left[t+\frac{t^{2}}{2}\right]_{0}^{2} = 4a \;\;\checkmark\]

\[\text{The centre moves as a point under constant acceleration } a \text{ with initial speed } a \;\;\checkmark\]