91Ó°ÊÓ

Velocity is the rate of change of displacement. It is vector quantity expressed as,

\(v = \frac{s}{t}\)

Here\(s\)is the displacement and\(t\)is the time.

The SI unit of velocity is meter per second\(\left( {{\rm{m/s}}} \right)\).

• The magnitude of the resultant velocity vector,\({v_{tot}} = 6.72\;{\rm{m/s}}\).
• Velocity vector\({{\rm{v}}_{\rm{A}}}\)makes an angle with horizontal is\({22.5^ \circ }\).
• The angle between resultant velocity and velocity vector\({{\rm{v}}_{\rm{A}}}\)is\({26.5^ \circ }\).
• The angle between resultant velocity and velocity vector \({{\rm{v}}_{\rm{B}}}\) is \({23.0^ \circ }\).

The two velocities \({{\rm{v}}_{\rm{A}}}\) and \({{\rm{v}}_{\rm{B}}}\) add to give a total \({{\rm{v}}_{{\rm{tot}}}}\) is represented as,

Two velocities \({{\rm{v}}_{\rm{A}}}\) and \({{\rm{v}}_{\rm{B}}}\) add to give a total \({{\rm{v}}_{{\rm{tot}}}}\)

The angle of between \(x\)-axis and \({{\rm{v}}_{{\rm{tot}}}}\) is,

\(\begin{aligned}{}\alpha &= \left( {{{26.5}^ \circ }} \right) + \left( {{{22.5}^ \circ }} \right)\\ &= {49^ \circ }\end{aligned}\)

The angle between the \(y\)-axis and \({{\rm{v}}_{\rm{B}}}\) is obtained from the triangle \(OAB\),

\(\begin{aligned}{}{180^ \circ } &= {22.5^ \circ } + {26.5^ \circ } + {23^ \circ } + \beta + {90^ \circ }\\\beta &= {180^ \circ } - {22.5^ \circ } - {26.5^ \circ } - {23^ \circ } - {90^ \circ }\\ &= {18^ \circ }\end{aligned}\)

The angle between the \(x\)-axis and \({{\rm{v}}_{\rm{B}}}\) is obtained from the triangle \(ABC\),

\(\begin{aligned}{}{180^ \circ } &= \beta + {90^ \circ } + \gamma \\\gamma &= {180^ \circ } - {90^ \circ } - \beta \end{aligned}\)

Substitute \({18^ \circ }\) for \(\beta \), and we get,

\(\begin{aligned}{}\gamma &= {180^ \circ } - {90^ \circ } - {18^ \circ }\\ &= {72^ \circ }\end{aligned}\)

The horizontal component of \({{\rm{v}}_{\rm{A}}}\) is,

\(\begin{aligned}{}{v_{{A_x}}} &= {v_A}\cos \left( {{{22.5}^ \circ }} \right)\\ &= 0.924{v_A}\end{aligned}\)

The horizontal component of \({{\rm{v}}_{\rm{B}}}\) is,

\({v_{{B_x}}} = {v_B}\cos \gamma \)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{v_{{B_x}}} &= {v_B}\cos \left( {{{72}^ \circ }} \right)\\ &= 0.309{v_B}\end{aligned}\)

The horizontal component of \({{\rm{v}}_{{\rm{tot}}}}\) is,

\({v_{to{t_x}}} = {v_{tot}}\cos \alpha \)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{v_{to{t_x}}} &= \left( {6.72\;{\rm{m/s}}} \right) \times \cos \left( {{{49}^ \circ }} \right)\\ &= 4.409\;{\rm{m/s}}\end{aligned}\)

The horizontal component of \({{\rm{v}}_{{\rm{tot}}}}\) is,

\({v_{to{t_x}}} = {v_{{A_x}}} + {v_{{B_x}}}\)

Substitute the values in the above expression, and we get,

\(4.409\;{\rm{m/s}} = 0.924{v_A} + 0.309{v_B}\)

The vertical component of \({{\rm{v}}_{\rm{A}}}\) is,

\(\begin{aligned}{}{v_{{A_y}}} &= {v_A}\sin \left( {{{22.5}^ \circ }} \right)\\ &= 0.385{v_A}\end{aligned}\)

The vertical component of \({{\rm{v}}_{\rm{B}}}\) is,

\({v_{{B_y}}} = {v_B}\cos \gamma \)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{v_{{B_y}}} &= {v_B}\sin \left( {{{72}^ \circ }} \right)\\ &= 0.951{v_B}\end{aligned}\)

The vertical component of \({{\rm{v}}_{{\rm{tot}}}}\) is,

\({v_{to{t_y}}} = {v_{tot}}\sin \alpha \)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{v_{to{t_y}}} &= \left( {6.72\;{\rm{m/s}}} \right) \times \sin \left( {{{49}^ \circ }} \right)\\ &= 5.072\;{\rm{m/s}}\end{aligned}\)

The vertical component of \({{\rm{v}}_{{\rm{tot}}}}\) is,

\({v_{to{t_y}}} = {v_{{A_y}}} + {v_{{B_y}}}\)

Substitute the values in the above expression, and we get,

\(5.072\;{\rm{m/s}} = 0.385{v_A} + 0.951{v_B}\)

Multiplying equation (1.1) with\(0.951\),

\(4.193\;{\rm{m/s}} = 0.879{v_A} + 0.294{v_B}\)

Multiplying equation (1.2) with\(0.309\),

\(1.567\;{\rm{m/s}} = 0.119{v_A} + 0.294{v_B}\)

Subtracting equation (1.4) from equation (1.3), we get,

\(\begin{aligned}{}\left( {4.193\;{\rm{m/s}}} \right) - \left( {1.567\;{\rm{m/s}}} \right) &= \left( {0.879{v_A} + 0.294{v_B}} \right) - \left( {0.119{v_A} + 0.294{v_B}} \right)\\2.626\;{\rm{m/s}} &= 0.76{v_A}\\{v_A} &= \frac{{2.626\;{\rm{m/s}}}}{{0.76}}\\ &= 3.455\;{\rm{m/s}}\end{aligned}\)

Substitute the values in the equation (1.3), and we get,

\(\begin{aligned}{}4.193\;{\rm{m/s}} &= 0.879 \times \left( {3.455\;{\rm{m/s}}} \right) + 0.294{v_B}\\0.294{v_B} &= 4.193\;{\rm{m/s}} - 0.879 \times \left( {3.455\;{\rm{m/s}}} \right)\\{v_B} &= \frac{{4.193\;{\rm{m/s}} - 0.879 \times \left( {3.455\;{\rm{m/s}}} \right)}}{{0.294}}\\ &= 3.932\;{\rm{m/s}}\end{aligned}\)

Hence, the magnitude of the velocities\({v_A}\)and\({v_B}\)are\(3.455\;{\rm{m/s}}\)and\(3.932\;{\rm{m/s}}\)respectively.