College Physics 3.15 – North and east components of a displacement


Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.59.

The figure shows a vector arrow starting from San Francisco going to Sacramento. The vector has a magnitude of 123 km and is directed 45° from the east direction.
Figure 3.59: The displacement from San Francisco to Sacramento

Solution:

North Component

The North component is given by

S_N=123\:km\:\cdot sin\left(45^{\circ} \right)=86.97\:km

East Component

The east component is

S_E=123\:km\:\cdot cos\:\left(45^{\circ} \right)=86.97\:km


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


College Physics 2.29 – Acceleration of freight trains


Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?


Solution:

Part A

The final velocity can be solved using the formula \text{v}=\text{v}_0+\text{at}. We substitute the given values.

\text{v}=\text{v}_0+\text{at}

\displaystyle \text{v}=4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right)

\text{v}=28.0\:\text{m/s}

Part B

Rearrange the equation we used in part (a) by solving in terms of t, we have

      \displaystyle \text{t}=\frac{\text{v}-\text{v}_0}{\text{a}}

      \displaystyle \text{t}=\frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2}

      \text{t}=50.91\:\sec

Part C

The change in position for part (a), \Delta \text{x}, or distance traveled is computed using the formula \Delta \text{x}=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2

     \Delta \text{x}=\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2

     \Delta \text{x}=7680\:\text{m}

For the situation in part (b), the distance traveled is computed using the formula \Delta \text{x}=\frac{\text{v}^2-\text{v}_0^2}{2\text{a}}

     \displaystyle \Delta \text{x}=\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)}

     \Delta \text{x}=712.73\:\text{m}