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Found in: Page 121

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Find the magnitudes of velocities $${v_A}$$ and $${v_B}$$ in Figure 3.57.The two velocities $${{\rm{v}}_{\rm{A}}}$$ and $${{\rm{v}}_{\rm{B}}}$$ add to give a total $${{\rm{v}}_{{\rm{tot}}}}$$ .

The magnitude of the velocities $${v_A}$$ and $${v_B}$$ are $$3.455\;{\rm{m/s}}$$ and $$3.932\;{\rm{m/s}}$$ respectively.

See the step by step solution

## Step 1: Definition of Velocity

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 $$\left( {{\rm{m/s}}} \right)$$.

## Step 2: Given data

• 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 $${23.0^ \circ }$$.

## Step 3: Finding angles

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}

## Step 4: Horizontal components of velocities

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,

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

## Step 5: Vertical components of velocities

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,

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

## Step 6: Solving equations

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.