The motion of a point on the circumference of a rolling wheel of radius 5 feet is described by the vector function
\[
\vec{r}(t)=5(24 t-\sin (24 t)) \vec{i}+5(1-\cos (24 t)) \vec{j}
\]
Find the velocity vector of the point.
\[
\vec{v}(t)=
\]
Find the acceleration vector of the point.
\[
\vec{a}(t)=
\]
Find the speed of the point.
\[
s(t)=
\]

Step 1 ：Given the position vector function of a point on the circumference of a rolling wheel of radius 5 feet as \(\vec{r}(t)=5(24 t-\sin (24 t)) \vec{i}+5(1-\cos (24 t)) \vec{j}\)

Step 2 ：The velocity vector of the point is the derivative of the position vector function, which is \(\vec{v}(t) = [120 - 120\cos(24t), 120\sin(24t)]\)

Step 3 ：The acceleration vector of the point is the derivative of the velocity vector function, which is \(\vec{a}(t) = [2880\sin(24t), 2880\cos(24t)]\)

Step 4 ：The speed of the point is the magnitude of the velocity vector, which is \(s(t) = \sqrt{(120 - 120\cos(24t))^2 + 14400\sin^2(24t)}\)

Step 5 ：Final Answer: The velocity vector of the point is \(\boxed{(120 - 120\cos(24t))\vec{i} + 120\sin(24t)\vec{j}}\)

Step 6 ：The acceleration vector of the point is \(\boxed{2880\sin(24t)\vec{i} + 2880\cos(24t)\vec{j}}\)

Step 7 ：The speed of the point is \(\boxed{\sqrt{(120 - 120\cos(24t))^2 + 14400\sin^2(24t)}}\)