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

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

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

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 $\boxed{{\displaystyle (120-120\cos (24t))\overrightarrow{i}+120\sin (24t)\overrightarrow{j}}}$

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

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