Step 1

We are given the speed of car $A$ as ${v_A} = 90\,{\rm{ft/s}}$, the increasing rate of car $A$’s speed as ${\dot v_A} = 15\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}$ the speed of car $B$ as ${v_B} = 105\,{\rm{ft/s}}$, the increasing rate of car $B$’s speed as ${\dot v_B} = 25\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}$, the radius of circular path of car $A$ as ${r_A} = 300\,{\rm{ft}}$, the radius of the circular path of car $B$ as ${r_B} = 250\,{\rm{ft}}$, and the angle position between both cars as $\theta = 60^\circ $.

We are asked to determine the relative velocity and relative acceleration of car $A$ with respect to car $B$.

 
Step 2

Find the Cartesian form of car $A$’s speed using the following relation.

\[{{\bf{v}}_A} = - {v_A}{\bf{i}}\]

On substituting the known value in the above equation we get,

\[{{\bf{v}}_A} = - \left( {90{\bf{i}}} \right)\,{\rm{ft/s}}\]
 
Step 3

Find the angle of velocity of car $B$ with the vertical using the following relation.

\[\alpha = 90^\circ - \theta \]

On substituting the known value in the above equation we get,

\[\begin{array}{c} \alpha = 90^\circ - 60^\circ \\ \alpha = 30^\circ \end{array}\]
 
Step 4

Find the Cartesian form of car $B$’s speed using the following relation.

\[{{\bf{v}}_B} = {v_B}\left( { - \sin \alpha {\bf{i}} + \cos \alpha {\bf{j}}} \right)\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {{\bf{v}}_B} = \left( {105\,{\rm{ft/s}}} \right)\left( { - \sin 30^\circ {\bf{i}} + \cos 30^\circ {\bf{j}}} \right)\\ {{\bf{v}}_B} = \left( { - 52.5{\bf{i}} + 90.93{\bf{j}}} \right)\,{\rm{ft/s}} \end{array}\]
 
Step 5

Consider the Cartesian form of relative velocity of car $A$ with respect to $B$ is ${{\bf{v}}_{A/B}} = {\left( {{v_{A/B}}} \right)_x}{\bf{i}} + {\left( {{v_{A/B}}} \right)_y}{\bf{j}}$.

Find the relative velocity of car $A$ with respect to car $B$ using the following relation.

\[{{\bf{v}}_{A/B}} = {{\bf{v}}_A} - {{\bf{v}}_B}\]

On substituting ${\left( {{v_{A/B}}} \right)_x}{\bf{i}} + {\left( {{v_{A/B}}} \right)_y}{\bf{j}}$ for ${{\bf{v}}_{A/B}}$ in the above equation we get,

\[{\left( {{v_{A/B}}} \right)_x}{\bf{i}} + {\left( {{v_{A/B}}} \right)_y}{\bf{j}} = {{\bf{v}}_A} - {{\bf{v}}_B}\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {\left( {{v_{A/B}}} \right)_x}{\bf{i}} + {\left( {{v_{A/B}}} \right)_y}{\bf{j}} = - \left( {90{\bf{i}}} \right)\,{\rm{ft/s}} - \left( { - 52.5{\bf{i}} + 90.93{\bf{j}}} \right)\,{\rm{ft/s}}\\ {\left( {{v_{A/B}}} \right)_x}{\bf{i}} + {\left( {{v_{A/B}}} \right)_y}{\bf{j}} = - \left( {37.5{\bf{i}}} \right)\,{\rm{ft/s}} - \left( {90.93{\bf{j}}} \right)\,{\rm{ft/s}} \end{array}\]
 
Step 6

On equating the coefficients of ${\bf{i}}$, we get,

\[{\left( {{v_{A/B}}} \right)_x} = - 37.5\,{\rm{ft/s}}\]

On equating the coefficients of ${\bf{j}}$, we get,

\[{\left( {{v_{A/B}}} \right)_y} = - 90.93\,{\rm{ft/s}}\]
 
Step 7

Find the magnitude of relative velocity of car $A$ with respect to $B$ using the following relation.

\[{v_{A/B}} = \sqrt {\left( {{v_{A/B}}} \right)_x^2 + \left( {{v_{A/B}}} \right)_y^2} \]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {v_{A/B}} = \sqrt {{{\left( { - 37.5\,{\rm{ft/s}}} \right)}^2} + {{\left( { - 90.93\,{\rm{ft/s}}} \right)}^2}} \\ {v_{A/B}} = 98.36\,{\rm{ft/s}} \end{array}\]
 
Step 8

The direction of the speed of car A with respect to car B, we get,

\[\begin{array}{c} \beta = {\tan ^{ - 1}}\left( {\frac{{{{\left( {{v_{A/B}}} \right)}_y}}}{{{{\left( {{v_{A/B}}} \right)}_x}}}} \right)\\ \beta = {\tan ^{ - 1}}\left( {\frac{{90.93\;{\rm{ft/s}}}}{{37.5\;{\rm{ft/s}}}}} \right)\\ \beta = 67.6^\circ \end{array}\]
 
Step 9

Find the normal acceleration of car $A$ using the following relation.

\[{a_{nA}} = \frac{{v_A^2}}{{{\rho _A}}}\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {a_{nA}} = \frac{{{{\left( {90\,{\rm{ft/s}}} \right)}^2}}}{{300\,{\rm{ft}}}}\\ {a_{nA}} = 27\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 10

Find the Cartesian form of the acceleration of car $A$ using the following relation.

\[{{\bf{a}}_A} = - {a_{nA}}{\bf{i}} - {\dot v_A}{\bf{j}}\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {{\bf{a}}_A} = - \left( {27\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right){\bf{j}} - \left( {15\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right){\bf{i}}\\ {{\bf{a}}_A} = \left( { - {\rm{15}}{\bf{i}} - {\rm{27}}{\bf{j}}} \right)\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 11

Find the normal acceleration of car $B$ using the following relation.

\[{a_{nB}} = \frac{{v_B^2}}{{{\rho _B}}}\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {a_{nB}} = \frac{{{{\left( {105\,{\rm{ft/s}}} \right)}^2}}}{{250\,{\rm{ft}}}}\\ {a_{nB}} = 44.1\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 12

Find the Cartesian form of the acceleration of car $B$ using the following relation.

\[{{\bf{a}}_B} = - {a_{nB}}{\bf{i}} + {\dot v_B}{\bf{j}}\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {{\bf{a}}_B} = - \left( {44.1\left( {\sin 60^\circ {\bf{i}} + \cos 60^\circ {\bf{j}}} \right)} \right)\,{\rm{ft/}}{{\rm{s}}^2} + 25\left( {\cos 60^\circ {\bf{i}} + \sin 60^\circ {\bf{j}}} \right)\,{\rm{ft/}}{{\rm{s}}^2}\\ {{\bf{a}}_B} = \left( { - 25.69{\bf{i}} - 43.7{\bf{j}}} \right){\rm{ft/}}{{\rm{s}}^2} \end{array}\]
 
Step 13

Consider the Cartesian form of relative acceleration of car $A$ with respect to $B$ is ${{\bf{a}}_{A/B}} = {\left( {{a_{A/B}}} \right)_x}{\bf{i}} + {\left( {{a_{A/B}}} \right)_y}{\bf{j}}$.

Find the relative acceleration of car $A$ with respect to $B$ using the following relation.

\[{{\bf{a}}_{A/B}} = {{\bf{a}}_A} - {{\bf{a}}_B}\]

On substituting ${\left( {{a_{A/B}}} \right)_x}{\bf{i}} + {\left( {{a_{A/B}}} \right)_y}{\bf{j}}$ for ${{\bf{a}}_{A/B}}$ in the above equation we get,

\[{\left( {{a_{A/B}}} \right)_x}{\bf{i}} + {\left( {{a_{A/B}}} \right)_y}{\bf{j}} = {{\bf{a}}_A} - {{\bf{a}}_B}\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {\left( {{a_{A/B}}} \right)_x}{\bf{i}} + {\left( {{a_{A/B}}} \right)_y}{\bf{j}} = \left( { - 15{\bf{i}} - 27{\bf{j}}} \right)\,{\rm{ft/}}{{\rm{s}}^2} - \left( { - 25.69{\bf{i}} - 43.7{\bf{j}}} \right)\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}\\ {\left( {{a_{A/B}}} \right)_x}{\bf{i}} + {\left( {{a_{A/B}}} \right)_y}{\bf{j}} = \left( {10.69{\bf{i}} + 16.7{\bf{j}}} \right)\,{\rm{ft/}}{{\rm{s}}^2} \end{array}\]
 
Step 14

On equating the coefficients of ${\bf{i}}$, we get,

\[{\left( {{a_{A/B}}} \right)_x} = 10.69\,{\rm{ft/}}{{\rm{s}}^2}\]

On equating the coefficients of ${\bf{j}}$, we get,

\[{\left( {{a_{A/B}}} \right)_y} = \left( {16.7} \right)\,{\rm{ft/}}{{\rm{s}}^2}\]
 
Step 15

Find the magnitude the relative acceleration of car $A$ with respect to $B$ using the following relation.

\[{a_{A/B}} = \sqrt {\left( {{a_{A/B}}} \right)_x^2 + \left( {{a_{A/B}}} \right)_y^2} \]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {a_{A/B}} = \sqrt {{{\left( {10.69\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right)}^2} + {{\left( {16.7\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}}} \right)}^2}} \\ {a_{A/B}} = 19.83\,{\rm{ft/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 16

Find the direction angle of the relative acceleration of car $A$ with respect to car $B$ using the following relation we get,

\[\gamma = {\tan ^{ - 1}}\left( {\frac{{{{\left( {{a_{A/B}}} \right)}_y}}}{{{{\left( {{a_{A/B}}} \right)}_x}}}} \right)\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} \gamma = {\tan ^{ - 1}}\left( {\frac{{16.7\,{\rm{ft/s}}}}{{10.69\,{\rm{ft/s}}}}} \right)\\ \gamma = 57.38^\circ \end{array}\]