Step 1

We are given the following data:

The value of $r$ is $r = 8\;{\rm{ft}}$.

The value of $\dot \theta $ is $\dot \theta = 2\;{\rm{rad/s}}$.

The equation of $z$ is $z = \left( {1.5\sin \theta } \right)\;{\rm{ft}}$.

The initial value of $\theta $ is $\theta = 0^\circ $. (At the starting of motion)

We are asked to determine the maximum and minimum magnitudes of the velocity and acceleration of the horse during the motion.

 
Step 2

We will differentiate the given equation of $r$ by using the chain rule on both sides with respect to $t$ to obtain $\dot r$.

\[\begin{array}{c} \frac{d}{{dt}}\left( r \right) = \frac{d}{{dt}}\left[ {8\;{\rm{ft}}} \right]\\ \dot r = 0\;{\rm{ft/s}} \end{array}\] … (1)
 
Step 3

Differentiate the equation (1) with respect to $t$ to obtain $\ddot r$.

\[\begin{array}{c} \frac{{d\dot r}}{{dt}} = \frac{d}{{dt}}\left( {0\;{\rm{ft/s}}} \right)\\ \ddot r = 0\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 4

Differentiate the given equation of $\dot \theta $ with respect to $t$ to obtain $\ddot \theta $.

\[\begin{array}{c} \frac{{d\dot \theta }}{{dt}} = \frac{d}{{dt}}\left[ {2\;{\rm{rad/s}}} \right]\\ \ddot \theta = 0\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 5

We will differentiate the given equation of $z$ by using the chain rule on both sides with respect to $\theta $ to obtain $\dot z$.

\[\begin{array}{c} \frac{d}{{d\theta }}\left( z \right) = \frac{d}{{d\theta }}\left[ {1.5\sin \theta \;{\rm{ft}}} \right]\\ \dot z = \left[ {\left( {1.5\cos \theta } \right)\left( {\dot \theta } \right)} \right]\;{\rm{ft/s}} \end{array}\] … (2)
 
Step 6

Differentiate the equation (2) with respect to $\theta $ to obtain $\ddot z$.

\[\begin{array}{c} \frac{{d\dot z}}{{dt}} = \frac{d}{{dt}}\left[ {\left( {1.5\cos \theta } \right)\left( {\dot \theta } \right)} \right]\;{\rm{ft/s}}\\ \ddot z = \left[ { - 1.5\sin \theta {{\left( {\dot \theta } \right)}^2} + 1.5\cos \theta \left( {\ddot \theta } \right)} \right]\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\] … (3)
 
Step 7

Substitute all the known values in the equation (2) to obtain the value of $\dot z$.

\[\begin{array}{c} \dot z = \left[ {\left( {1.5\cos \left( {0^\circ } \right)} \right)\left( {2\;{\rm{rad/s}}} \right)} \right]\;{\rm{ft/s}}\\ = 3\;{\rm{ft/s}} \end{array}\]
 
Step 8

Substitute all the known values in the equation (3) to obtain the value of $\ddot z$.

\[\begin{array}{c} \ddot z = \left[ { - 1.5\sin \left( {0^\circ } \right){{\left( {2\;{\rm{rad/s}}} \right)}^2} + 1.5\cos \left( {0^\circ } \right)\left( 0 \right)} \right]\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}\\ = 0\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 9

The formula to calculate the radial velocity ${v_r}$ is given by,

\[{v_r} = \dot r\]

Substitute all the known values in the above formula.

\[{v_r} = 0\;{\rm{ft/s}}\]
 
Step 10

The formula to calculate the transverse velocity ${v_\theta }$ is given by,

\[{v_\theta } = r\dot \theta \]

Substitute all the known values in the above formula.

\[\begin{array}{c} {v_\theta } = \left[ {\left( {8\;{\rm{ft}}} \right)\left( {2\;{\rm{rad/s}}} \right)} \right]\;{\rm{ft/s}}\\ {\rm{ = }}\;16\;{\rm{ft/s}} \end{array}\]
 
Step 11

The formula to calculate the velocity ${\left( {{v_z}} \right)_{\max }}$ is given by,

\[{\left( {{v_z}} \right)_{\max }} = \dot z\]

Substitute all the known values in the above formula.

\[{\left( {{v_z}} \right)_{\max }} = 3\;{\rm{ft/s}}\]
 
Step 12

For minimum value of ${v_z}$, the value of $\theta $ sould be $90^\circ $ means $\theta = 90^\circ $.

To calculate the velocity ${\left( {{v_z}} \right)_{\min }}$, we need to substitute the value of $\theta $ in equation (2).

\[\begin{array}{c} {\left( {{v_z}} \right)_{\min }} = \left[ {1.5\cos \left( {90^\circ } \right)\left( {2\;{\rm{rad/s}}} \right)} \right]\;{\rm{ft/s}}\\ {\rm{ = }}\left[ {1.5\left( 0 \right)\left( {2\;{\rm{rad/s}}} \right)} \right]\;{\rm{ft/s}}\\ {\rm{ = }}\;{\rm{0}}\;{\rm{ft/s}} \end{array}\]
 
Step 13

The formula to calculate the magnitude of the maximum velocity is given by,

\[{v_{\max }} = \sqrt {{{\left( {{v_\theta }} \right)}^2} + {{\left( {{v_z}} \right)}^2}_{\max }} \]

Substitute all the known values in the above formula.

\[\begin{array}{c} {v_{\max }} = \sqrt {{{\left( {16\;{\rm{ft/s}}} \right)}^2} + {{\left( {3\;{\rm{ft/s}}} \right)}^2}} \\ = \sqrt {265\;{\rm{f}}{{\rm{t}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\ = 16.3\;{\rm{ft/s}} \end{array}\]
 
Step 14

The formula to calculate the magnitude of the minimum velocity is given by,

\[{v_{\min }} = \sqrt {{{\left( {{v_\theta }} \right)}^2} + {{\left( {{v_z}} \right)}^2}_{\min }} \]

Substitute all the known values in the above formula.

\[\begin{array}{c} {v_{\min }} = \sqrt {{{\left( {16\;{\rm{ft/s}}} \right)}^2} + {{\left( {0\;{\rm{ft/s}}} \right)}^2}} \\ = 16\;{\rm{ft/s}} \end{array}\]
 
Step 15

The formula to calculate the radial component of the acceleration is given by,

\[{a_r} = \left[ {\ddot r - r{{\left( {\dot \theta } \right)}^2}} \right]\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {a_r} = \left[ {\left( 0 \right) - \left( {8\;{\rm{ft}}} \right){{\left( {2\;{\rm{rad/s}}} \right)}^2}} \right]\\ = - 32\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 16

The formula to calculate the transverse component of the acceleration is given by,

\[{a_\theta } = \left( {r\ddot \theta + 2\dot r\dot \theta } \right)\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {a_\theta } = \left[ {\left( {8\;{\rm{ft}}} \right)\left( 0 \right) + 2\left( 0 \right)\left( {2\;{\rm{rad/s}}} \right)} \right]\;\\ = 0\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 17

For maximum value of ${a_z}$, the value of $\theta $ sould be $90^\circ $ means $\theta = 90^\circ $.

To calculate the maximum acceleration ${\left( {{a_z}} \right)_{\max }}$, we need to substitute the value of $\theta $ in equation (3).

\[\begin{array}{c} {\left( {{a_z}} \right)_{\max }} = \left[ { - 1.5\sin 90^\circ {{\left( {2\;{\rm{rad/s}}} \right)}^2} + 1.5\cos 90^\circ \left( 0 \right)} \right]\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}\\ = \left[ { - 6 + 0} \right]\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}\\ = - 6\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 18

For minimum value of ${a_z}$, the value of $\theta $ sould be $0^\circ $ means $\theta = 0^\circ $.

To calculate the minimum acceleration ${\left( {{a_z}} \right)_{\min }}$, we need to substitute the value of $\theta $ in equation (3).

\[\begin{array}{c} {\left( {{a_z}} \right)_{\min }} = \left[ { - 1.5\sin 0^\circ {{\left( {2\;{\rm{rad/s}}} \right)}^2} + 1.5\cos 0^\circ \left( 0 \right)} \right]\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}\\ = \left[ {0 + 0} \right]\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}\\ = 0\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 19

The formula to calculate the magnitude of the maximum acceleration is given by,

\[{a_{\max }} = \sqrt {{{\left( {{a_r}} \right)}^2} + {{\left( {{a_\theta }} \right)}^2}_{\max } + {{\left( {{a_z}} \right)}^2}_{\max }} \]

Substitute all the known values in the above formula.

\[\begin{array}{c} {a_{\max }} = \sqrt {{{\left( { - 32\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( { - 6\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right)}^2}} \\ = 32.6\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 20

The formula to calculate the magnitude of the minimum acceleration is given by,

\[{a_{\min }} = \sqrt {{{\left( {{a_r}} \right)}^2} + {{\left( {{a_\theta }} \right)}^2}_{\min } + {{\left( {{a_z}} \right)}^2}_{\min }} \]

Substitute all the known values in the above formula.

\[\begin{array}{c} {a_{\min }} = \sqrt {{{\left( { - 32\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( {0\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right)}^2}} \\ = 32\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]