Chapter 3, Problem 49P

(a)

To determine

To Calculate: The magnitude and the direction of $\Delta {\overrightarrow{v}}_{av}$ during the $3.0\text{ s}$ interval.

Answer to Problem 49P

Magnitude of the resultant velocity vector $\Delta {\overrightarrow{v}}_{av}$ is $36.06{\text{ ms}}^{\text{-1}}.$

Direction of the resultant velocity vector $\Delta {\overrightarrow{v}}_{av}$ is ${47.15}^{\circ }$ .

Explanation of Solution

Given data

Velocity at $\text{t = 0 s}$ = $40\text{ m}{\text{.s}}^{\text{-1}}$

The angle of velocity at $\text{t = 0 s}$ = $\theta ={45}^{\circ }$

Velocity at $\text{t = 3 s =}$ $30\text{ m}{\text{.s}}^{\text{-1}}$

The angle of velocity at $\text{t =}\text{3 s =}$$\theta ={50}^{\circ }$

Displacement of x at $\text{t =}\text{3 s}$$=100\text{m}$

Displacement of y at $\text{t =}\text{3 s}$ = $80\text{m}$

Formula used

  

  $\begin{array}{l}\left|\overrightarrow{v}\right|=\sqrt{{\left|{\overrightarrow{v}}_x\right|}^2+{\left|{\overrightarrow{v}}_y\right|}^2}\\ \theta ={\tan }^{-1}\left(\frac{\left|{\overrightarrow{v}}_y\right|}{\left|{\overrightarrow{v}}_x\right|}\right)\end{array}$

  $\left|v\right|$ denotes the magnitude of the vector and $\theta$ denotes the direction of the vector v

  $a=\frac{\Delta v}{\Delta t}$

where a is the acceleration and v and t represent velocity and time respectively.

Calculation

At $t=0\text{ s}$ .

Velocity component in horizontal direction = $40\text{ m}{\text{.s}}^{\text{-1}}\times \cos {45}^{\circ }=40\text{ m}{\text{.s}}^{\text{-1}}\times \frac{1}{\sqrt{2}}=28.284\text{ m}{\text{.s}}^{\text{-1}}$

Velocity component in vertical direction = $40\text{ m}{\text{.s}}^{\text{-1}}\times \sin {45}^{\circ }=40\text{ m}{\text{.s}}^{\text{-1}}\times \frac{1}{\sqrt{2}}=28.284\text{ m}{\text{.s}}^{\text{-1}}$

At $t=3.0\text{ s}$ ,

Velocity component in horizontal direction = $30\text{ m}{\text{.s}}^{\text{-1}}\times \cos {50}^{\circ }=30\text{ m}{\text{.s}}^{\text{-1}}\times 0.6428=19.284\text{ m}{\text{.s}}^{\text{-1}}$

Velocity component in vertical direction = $30\text{ m}{\text{.s}}^{\text{-1}}\times \sin {50}^{\circ }=30\text{ m}{\text{.s}}^{\text{-1}}\times 0.7660=22.98\text{ m}{\text{.s}}^{\text{-1}}$

Total velocity components in horizontal direction = $28.284\text{ m}{\text{.s}}^{\text{-1}}+19.284\text{ m}{\text{.s}}^{\text{-1}}=47.568\text{ m}{\text{.s}}^{\text{-1}}$

Total velocity components in vertical direction = $28.284\text{ m}{\text{.s}}^{\text{-1}}+22.98\text{ m}{\text{.s}}^{\text{-1}}=51.264\text{ m}{\text{.s}}^{\text{-1}}$

Resultant velocity after $t=3\text{ s}$ = $\sqrt{{47.568}^2+{51.264}^2}\text{ m}{\text{.s}}^{\text{-1}}=69.93\text{ m}{\text{.s}}^{\text{-1}}$

  $\theta ={\tan }^{-1}\left(\frac{51.264}{47.568}\right)={\tan }^{-1}\left(1.078\right)={47.15}^{\circ }$

Conclusion

The magnitude of the resultant velocity vector $\Delta {\overrightarrow{v}}_{av}$ = $36.06{\text{ ms}}^{\text{-1}}$

The direction of the resultant velocity vector $\Delta {\overrightarrow{v}}_{av}$ = ${47.15}^{\circ }$ from the horizontal direction to vertical direction.

(b)

To determine

To Calculate: The magnitude and the direction of ${\overrightarrow{a}}_{av}$ .

Answer to Problem 49P

Magnitude of the resultant acceleration vector ${\overrightarrow{a}}_{av}$ is $3.482{\text{ms}}^{\text{-2}}$

Direction of the resultant acceleration vector ${\overrightarrow{a}}_{av}$ is. ${30.50}^{\circ }$ from negative x direction to negative y direction

Explanation of Solution

Given data

Velocity at $\text{t = 0 s}$ = $40\text{ m}{\text{.s}}^{\text{-1}}$

The angle of velocity at $\text{t = 0 s}$ = $\theta ={45}^{\circ }$

Velocity at $\text{t = 3 s =}$$30\text{ m}{\text{.s}}^{\text{-1}}$

The angle of velocity at $\text{t =}\text{3 s =}$ $\theta ={50}^{\circ }$

Displacement of x at $\text{t =}\text{3 s}$ $=100\text{m}$

Displacement of y at $\text{t =}\text{3 s}$ = $80\text{m}$

Formula used

  $\overrightarrow{a}=\frac{\Delta \overrightarrow{v}}{\Delta t}$

  $\begin{array}{l}\left|{\overrightarrow{a}}_{av}\right|=\sqrt{{\left|{\overrightarrow{a}}_x\right|}^2+{\left|{\overrightarrow{a}}_y\right|}^2}\\ \theta ={\tan }^{-1}\left(\frac{\left|{\overrightarrow{a}}_y\right|}{\left|{\overrightarrow{a}}_x\right|}\right)\end{array}$

Calculation

  $\begin{array}{l}\overrightarrow{a}=\frac{\Delta \overrightarrow{v}}{\Delta t}\\ {\overrightarrow{a}}_x=\frac{19.284-28.284}{3}{\text{ ms}}^{\text{-2}}\\ {\overrightarrow{a}}_x=-\frac{9}{3}{\text{ ms}}^{\text{-2}}\\ {\overrightarrow{a}}_x=-3{\text{ ms}}^{\text{-2}}\\ {\overrightarrow{a}}_y=\frac{22.98-28.284}{3}{\text{ ms}}^{\text{-2}}\\ {\overrightarrow{a}}_y=-\frac{5.304}{3}{\text{ ms}}^{\text{-2}}\\ {\overrightarrow{a}}_y=-1.768{\text{ ms}}^{\text{-2}}\end{array}$

  $\begin{array}{l}\left|{\overrightarrow{a}}_{av}\right|=\sqrt{{\left|{\overrightarrow{a}}_x\right|}^2+{\left|{\overrightarrow{a}}_y\right|}^2}\\ \left|{\overrightarrow{a}}_{av}\right|=\sqrt{{\left(-3\right)}^2+{\left(-1.768\right)}^2}{\text{ms}}^{\text{-2}}\\ \left|{\overrightarrow{a}}_{av}\right|=\sqrt{9+3.126}{\text{ms}}^{\text{-2}}\\ \left|{\overrightarrow{a}}_{av}\right|=\sqrt{12.126}{\text{ms}}^{\text{-2}}\\ \left|{\overrightarrow{a}}_{av}\right|=3.482{\text{ms}}^{\text{-2}}\\ \theta ={\tan }^{-1}\left(\frac{\left|{\overrightarrow{a}}_y\right|}{\left|{\overrightarrow{a}}_x\right|}\right)\\ \theta ={\tan }^{-1}\left(\frac{-1.768}{-3}\right)\\ \theta ={\tan }^{-1}\left(0.589\right)\\ \theta ={30.50}^{\circ }\end{array}$

The magnitude of the resultant acceleration vector ${\overrightarrow{a}}_{av}$ = $3.482{\text{ms}}^{\text{-2}}$

The direction of the resultant acceleration vector ${\overrightarrow{a}}_{av}$ = ${30.50}^{\circ }$ from negative x-direction to negative y-direction

Conclusion

Magnitude of the resultant acceleration vector ${\overrightarrow{a}}_{av}$ is $3.482{\text{ms}}^{\text{-2}}$

Direction of the resultant acceleration vector ${\overrightarrow{a}}_{av}$ is ${30.50}^{\circ }$ from negative x direction to negative y direction.