College Physics 3.7 – Vector Subtraction


(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east (which is equivalent to subtracting B from A —that is, to finding R′ =A−B ).

(b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º south of west and then 12.0 m in a direction 20.0º east of south (which is equivalent to subtracting A from B —that is, to finding R′′ = B – A = – R′ ). Show that this is the case.


Solution:

Part A

We are required to solve for 

\vec{R'}=\vec{A'}+\vec{B}'

The x-component of the resultant is:

R_x'=A_x'+B_x'

R_x'=12\:cos\left(110^{\circ} \right)+\left(20.0\right)cos\left(40^{\circ} \right)=11.217\:m

The y-component of the resultant is:

R_x'=A_y'+B_y'

R_x'=\left(12.0\right)sin\left(110^{\circ} \right)+\left(20.0\right)sin\left(40^{\circ} \right)=24.132\:m

The resultant is

R'=\sqrt{R_x'+R_y'}

R'=\sqrt{\left(11.217\:m\right)^2+\left(24.132\:m\right)^2}

R'=26.6\:m

The compass direction is

\alpha =tan^{-1}\left(\frac{R_y'}{R_x'}\right)

\alpha =tan^{-1}\left(\frac{24.132}{11.217}\right)=65.1^{\circ}

The compass direction is 65.1^{\circ} \:North\:of\:East

Part B

The result is the same as that in Part A since the components did not change. Only the order has changed. This is consistent with the fact that \vec{A}-\vec{B}=-\left(\vec{B}-\vec{A}\right)


College Physics 3.6 – The resultant of two vectors added in reversed order


Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B , which is 20.0 m in a direction exactly 40º south of west, and then leg A , which is 12.0 m in a direction exactly 20º west of north. (This problem shows that A+B=B+A.)


Solution:

Reversing the order of the vectors, we come up with the triangle as shown below.

Two vectors A and B are added in reversed order to produce the resultant R with magnitude R and directed θ degrees from the horizontal

We can actually see that the resultant can be solved by considering the triangle formed by the three vectors. The same with the procedures used in the previous problem, we have

\text{R}^2=\text{A}^2+\text{B}^2-2\text{AB}\:\cos 70^{\circ}

\text{R}=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\:\text{cos}\:70^{\circ} }\:

\text{R}=\sqrt{\text{(12 m)}^2+\text{(20 m)}^2-2\text{(12 m)(20 m)}\:\text{cos}\:70^{\circ} }\:

\text{R}=19.4892\:\text{m}

To solve for the direction, we can solve for the angle included between B and R first.

\displaystyle \frac{12\:\text{m}}{\sin \text{C}}=\frac{19.4892\:\text{m}}{\sin 70^{\circ} }

\displaystyle \text{C}=\sin ^{-1}\left(\frac{12\:\text{m}\:\left(\sin 70^{\circ} \right)}{19.4892\:\text{m}}\right)

\text{C}=35.35^{\circ}

Finally to solve for the direction of the resultant, we have

\theta =40^{\circ} -35.35^{\circ} =4.65^{\circ}

The compass direction is 4.65^{\circ} \:\text{South of West}.

This result is actually the same as in the previous problem when the order of the vectors are not reversed. No matter what is the order of the vectors we are adding, the result is still the same. This proves that \vec{A}+\vec{B}=\vec{B}+\vec{A}.


College Physics 3.5 – Resultant of two vectors


Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure 3.54, then this problem finds their sum R=A+B.)

The two displacements A and B are added to give a total displacement of R with magnitude R and direction θ measured from the x-axis
Figure 3.54. The two displacements A and B are added to give a total displacement of R with magnitude R and direction θ measured from the x-axis

Solution:

To solve for the magnitude and direction of the resultant R, we can refer to the triangle formed by the tree vectors as shown in the figure below.

Two vectors A and B are given and added to find for the magnitude and direction of the resultant vector R

From the figure above, we are given the two sides A and B, and the included angle measured to be 70°. To solve for the magnitude of R, we can use the cosine law.

\text{R}^2=\text{A}^2+\text{B}^2-2\text{AB}\:\cos 70^{\circ}

\text{R}=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\:\text{cos}\:70^{\circ} }\:

\text{R}=\sqrt{\text{(12 m)}^2+\text{(20 m)}^2-2\text{(12 m)(20 m)}\:\text{cos}\:70^{\circ} }\:

\text{R}=19.4892\:\text{m}

To solve for the direction of the resultant, we just need to solve for angle C, then subtract 70° to get the value of θ. We can solve for angle C using the sine law.

\displaystyle \frac{20\:\text{m}}{\sin \text{C}}=\frac{19.4892\:\text{m}}{\sin 70^{\circ} }

\displaystyle \sin \:\text{C}=\frac{20\:\text{m}\:\left(\sin \:\:70^{\circ \:}\right)}{19.4892\:\text{m}}

\displaystyle \text{C}=\text{sin}^{-1}\left(\frac{20\:\text{m}\:\left(\sin \:\:70^{\circ \:}\right)}{19.4892\:\text{m}}\right)

\text{C}=74.65^{\circ}

Finally, to solve for θ, we subtract 70° from 74.65°.

\theta =74.65^{\circ} -70^{\circ} =4.65^{\circ}

Therefore, the compass direction of the resultant displacement is 4.65° South of West.


College Physics 3.4 – Resultant of west and north displacements


Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.53, then this problem asks you to find their sum R = A + B .)

The two displacements A and B add to give a total displacement R having magnitude R and direction θ.
Figure 3.53. The two displacements A and B add to give a total displacement R having magnitude R and direction θ.

Solution:

Refer to the diagram

3.4.PNG

The magnitude of the displacement (distance from the starting point to the final position) is given by 

R=\sqrt{A^2+B^2}

R=\sqrt{\left(18.0\:m\right)^2+\left(25.0\:m\right)^2}

R=30.8\:m

The compass direction of a line connecting your starting point to your final position is given by

\displaystyle \theta =tan^{-1}\left(\frac{B}{A}\right)

\displaystyle \theta =tan^{-1}\left(\frac{25.0\:m}{18.0\:m}\right)

\displaystyle \theta =54.25^{\circ}

The compass reading is 54.25^{\circ} ,\:\text{North of West}.


College Physics 3.3 – North and East components of a displacement


Find the north and east components of the displacement for the hikers shown in Figure 3.50.

A displacement S is shown with magnitude of 5.0 km and directed 40 degrees north of east.
Figure 3.50

Solution:

To solve for the north and east components of the given displacement vector, we need to draw them first by virtue of the parallelogram law. The vector, together with its components, is shown in the figure below.

A displacement vector S with magnitude of 5.0 km and directed at 40° north of east. Also shows the east and west component of the vector using the parallelogram law.

We can solve the north and east components by considering the right triangle ABC. The east component is the side AC and the north component is the side BC.

The North-component is given by

\displaystyle \text{s}_{\text{north}}=\left(5.0\:\text{km}\right)\text{sin}\:\left(40.0\right)^{\circ}

\text{s}_{\text{north}}=3.21\:\text{km}

The East-component is given by

\displaystyle \text{s}_{\text{east}}=\left(5.0\:\text{km}\right)\text{cos}\left(40.0\right)^{\circ}

\text{s}_{\text{east}}=3.83\:\text{km}


College Physics 3.2 – Distance vs displacement


Find the following for path B in Figure 3.52:

(a) the total distance traveled, and

(b) the magnitude and direction of the displacement from start to finish.

Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Solution:

Part A

So, the total distance traveled on path B as shown by the red line is just the sum of the distances. 

d=\left(4\times 120\:m\right)+\left(3\times 120\:m\right)+\left(3\times 120\:m\right)

d=1200\:m

Part B

The displacement is just the shortest distance from the beginning to the end of the path. The displacement has a magnitude given by

s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}

s=\sqrt{\left(1\times 120\:m\right)^2+\left(3\times 120\:m\right)^2}

s=379\:m

The direction of the displacement is given by

\theta =tan^{-1}\left(\frac{1\times 120\:m}{3\times 120\:m}\right)=18.4^{\circ} ,\:East\:of\:North


College Physics 3.1 – Distance and displacement computation


Find the following for path A in Figure 3.52:

(a) the total distance traveled, and

(b) the magnitude and direction of the displacement from start to finish.

Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Solution:

Part A

So, the total distance traveled on path A as shown by the green line is just the sum of the distances. 

d=\left(120\:m\times 3\right)+\left(120\:m\times 1\right)

d=480\:m

Part B

The displacement is just the shortest distance from the beginning to the end of the path. The displacement has a magnitude given by

s=\sqrt{\left(1\times 120\:m\right)^2+\left(3\times 120\:m\right)^2}=379\:m

The direction of the displacement is given by

\theta =tan^{-1}\left(\frac{1\times 120\:m}{3\times 120\:m}\right)=18.4^{\circ} ,\:East\:of\:North


Skateboarder on a ramp| Physics

A skateboarder starts up a 1.0-m-high, 30° ramp at a speed of 6.9 m/s. The skateboard wheels roll without friction. At the top, she leaves the ramp and sails through the air.

A) How far from the end of the ramp does the skateboarder touch down?

Continue reading “Skateboarder on a ramp| Physics”

Angular Acceleration| Circular Motion| Physics

Your car tire is rotating at 4.0 rev/s when suddenly you press down hard on the accelerator. After traveling 300 m, the tire’s rotation has increased to 6.5 rev/s . The radius of the tire is 32 cm.

A) What was the tire’s angular acceleration? Give your answer in rad/s²?

Continue reading “Angular Acceleration| Circular Motion| Physics”

Revolutions from angular velocity vs time graph| Circular Motion| Physics


The figure shows the angular-velocity-versus-time graph for a particle moving in a circle.

How many revolutions does the object make during the first 4.0 s?


Solution:

Part A

The total number of radians made by the object is the area under the graph. Based on the given graph, the total number of radians is 

\displaystyle \theta _{rad}=60\:radians

Convert this to revolutions, knowing that 2π radians is equal to 1 revolution. 

\displaystyle  revolutions=60\:rad\times \frac{1\:rev}{2\pi }=\frac{60}{2\pi }rev=9.5\:rev