Tag Archives: two-dimensional kinematics

College Physics by Openstax Chapter 3 Problem 5

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.)

Figure 3.54

Solution:

Consider Figure 3.5A shown below.

Figure 3.5A

Before we can use cosine law to solve for the magnitude of R, we need to solve for the interior angle 𝛽 first. The value of 𝛽 can be calculated by inspecting the figure and use simple knowledge on geometry. It is equal to the sum of 20° and the complement of 40°. That is

\beta = 20^\circ +\left( 90^\circ -40^\circ  \right) = 70^\circ

We can use cosine law to solve for R.

\begin{align*}
R^2 & =A^2+B^2 -2AB \cos \beta \\
R^2 & = \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right)
 \cos 70^\circ \\
R & = \sqrt{ \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right)
 \cos 70^\circ} \\
R  & =19.4892 \ \text{m} \\
R & =19.5 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

We can solve for α using sine law.

\begin{align*}
\frac{\sin \alpha}{B} & = \frac{\sin \beta}{R} \\
\frac{\sin \alpha}{20.0\ \text{m}} & = \frac{\sin 70^\circ }{19.4892 \ \text{m}} \\
\sin \alpha & = \frac{20.0 \ \sin 70^\circ }{19.4892} \\
\alpha & = \sin ^{-1}  \left(  \frac{20.0 \ \sin 70^\circ }{19.4892}  \right) \\
\alpha & = 74.6488 ^\circ 
\end{align*}

Then we solve for the value of θ by subtracting 70° from α.

\theta=74.6488 ^\circ -70 ^\circ = 4.65^\circ

Therefore, the compass reading is

4.65^\circ, \text{South of West} \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}
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College Physics by Openstax Chapter 3 Problem 4

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 \vec{A} and \vec{B} , as in Figure 3.53, then this problem asks you to find their sum \vec{R}=\vec{A}+\vec{B} .)

Figure 3.53

Solution:

Figure 3.4A

Consider Figure 3.54A.

The resultant of the two vectors \vec{A} and \vec{B} is labeled \vec{R}. This \vec{R} is directed \theta ^{\circ} from the x-axis.

We shall use the right triangle formed to solve for the unknowns.

Solve for the magnitude of the resultant.

\begin{align*}
R & = \sqrt{A^2 +B^2} \\
R & = \sqrt{\left(18.0 \ \text{m}  \right)^2+\left( 25.0 \ \text{m} \right)^2} \\
R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Solve for the value of \theta .

\begin{align*}
\theta & = \arctan \left( \frac{B}{A} \right) \\
\theta & = \arctan \left( \frac{25.0 \ \text{m}}{18.0 \ \text{m}} \right) \\
\theta & = 54.2^\circ 
\end{align*}

We need the complementary angle for the compass angle. 

\begin{align*}
90^\circ -54.2^\circ =35.8^\circ 
\end{align*}

Therefore, the compass angle reading is

\begin{align*}
35.8^\circ , \text{W of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
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College Physics by Openstax Chapter 3 Problem 3

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

Figure 3.50

Solution:

Refer to Figure 3-3-A for the north and east components of the displacement s of the hikers.

Figure 3-3-A

Considering the right triangle formed. The north component is computed as

\begin{align*}
\text{s}_{\text{north}} & = \left( 5.00 \ \text{km}  \right)\sin 40^\circ  \\
\text{s}_{\text{north}} & = 3.21 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}

Using the same right triangle, the east component is computed as follows. 

\begin{align*}
\text{s}_{\text{east}} & = \left( 5.00 \ \text{km}  \right)\cos 40^\circ  \\
\text{s}_{\text{east}} & = 3.83 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
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College Physics by Openstax Chapter 3 Problem 2

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 3.54 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

The total distance traveled is 

\begin{align*}

\text{d} & = \left(4 \times 120 \text{m} \right) + \left(3 \times 120\ \text{m} \right) + \left(3 \times 120\ \text{m} \right) \\
\text{d} & = 1 200\ \text{m}  \ \qquad \  {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part B

The magnitude of the displacement is 

\begin{align*}
\text{s} & = \sqrt{\left( s_x \right)^2+\left( s_y \right)^2} \\
\text{s} & = \sqrt{\left( 1 \times 120\ \text{m} \right)^2+ \left( 3 \times 120 \ \text{m} \right)^2} \\
\text{s} & = \sqrt{\left( 120\ \text{m} \right)^2+ \left( 360 \ \text{m} \right)^2} \\
\text{s} & = 379 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

The direction is

\begin{align*}
\theta & = \arctan \left( \frac{s_y}{s_x} \right) \\
\theta & = \arctan \left( \frac{360\ \text{m}}{120\ \text{m}} \right) \\
\theta & = 71.6^\circ , \ \text{N of E} \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
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College Physics by Openstax Chapter 2 Problem 57

A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

Solution:

Part A

Figure A

Consider Figure A.

We are interested in two positions. Position 1 is where the coin is dropped. At this position, the coin is 300 m above the ground, the time is 0 s, and the velocity is 10.0 m/s upward.

Position 2 is the highest point of the coin reaches. At this position, the velocity is equal to 0 m/s.

Position 1 is the initial position and position 2 is the final position. Solve for the value of y2.

\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 +2a \Delta y \\
\Delta y & = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} \\
y_2 - y_1 & =  \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} \\
y_2& =  \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} +y_1 \\
y_2 & = \frac{\left( 0 \ \text{m/s} \right)^2-\left( 10.0\ \text{m/s} \right)^2}{2\left( -9.81 \ \text{m/s}^2 \right)}+300\ \text{m}
\\
y_2 & =305 \ \text{m} \ \qquad  {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}

\therefore The maximum height reached by the coin is about 305 meters from the ground.

Part B

We do not know the position 4 seconds after the coin has been released, the answer can be above or below the initial point. We can actually use one of the kinematical equations to solve for the final position given the time. Here, the initial position is the point of release and the final position is the point of interest at 4.00 seconds after release.

\begin{align*}
\Delta y & = v_{y_1}t+\frac{1}{2}at^2 \\
y_2 - y_1 & = v_{y_1}t+\frac{1}{2}at^2 \\
y_2 & = y_1 +  v_{y_1}t+\frac{1}{2}at^2 \\
y_2 & = 300\ \text{m} +\left( 10\ \text{m/s} \right)\left( 4.00\ \text{s} \right)+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 4.00\ \text{s} \right)^2\\
y_2 & = 261.52\ \text{m} \\
y_2 & = 262\ \text{m}\ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}  \\
\end{align*}

\therefore The coin is at a height of 262 meters above the ground 4.00 seconds after release. That is, the coin is already dropping and it is already below the release point.

Solving for the velocity 4.00 seconds after release considering the same initial and final position.

\begin{align*}
v_{y_2} & = v_{y_1}+at \\
v_{y_2} & = 10\ \text{m/s} + \left( -9.81\ \text{m/s}^2 \right)\left( 4.00 \ \text{s} \right) \\
v_{y_2} & = -29.2\ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

\therefore The coin has a velocity of 29.2 m/s directed downward 4.00 seconds after it is released. This confirms that the coin is indeed moving downwards at this point.

Part C

Figure C

Considering figure C, we have two positions. Position 1 is the point of release 300 m above the ground with a velocity of 10 m/s upward. This is time 0 s.

The second position is at the ground where y=0 m. We are interested at the time in this position.

Considering position 1 as the initial position and position 2 as the final position.

\begin{align*}
\Delta y & = v_{y_1} \Delta t+\frac{1}{2}a\left( \Delta t \right)^2 \\
y_2-y_1 & =  v_{y_1}t+\frac{1}{2}at^2 \\
0\ \text{m}-300\ \text{m} & = \left( 10 \ \text{m/s} \right)t+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)t^2 \\
-300 & = 10t-4.905t^2 \\
4.905t^2-10t-300 & = 0 \\
\end{align*}

Solve for the value of t using the quadratic formula with a=4.905, b=-10, and c=-300.

\begin{align*}
t & = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\\
t & = \frac{-\left( -10 \right) \pm \sqrt{\left( 10 \right)^2-4\left( 4.905 \right)\left( -300 \right)}}{2\left( 4.905 \right)}\\
t & = 8.91 \ \text{s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

\therefore The time is about 8.91 seconds before the coin hits the ground.

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College Physics by Openstax Chapter 3 Problem 32

Verify the ranges shown for the projectiles in Figure 3.40(b) for an initial velocity of 50 m/s at the given initial angles.

Solution:

To verify the given values in the figure, we need to solve for individual ranges for the given initial angles. To do this, we shall use the formula

\displaystyle \text{R}=\frac{\text{v}_{\text{0}}^2 \sin 2\theta _{\text{0}}}{\text{g}}

When the initial angle is 15°, the range is

\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 15^{\circ} \right)}{9.81\:\text{m/s}^2}=127.42\:\text{m}

When the initial angle is 45°, the range is

\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=254.84\:\text{m}

When the initial angle is 75°, the range is

\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 75^{\circ} \right)}{9.81\:\text{m/s}^2}=127.42\:\text{m}

Based on the result of the calculations, we can say that the numbers in the figure are verified. The very small differences are only due to round-off errors.

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College Physics by Openstax Chapter 3 Problem 31

Verify the ranges for the projectiles in Figure 3.40(a) for θ=45º and the given initial velocities.

Solution:

To verify the given values in the figure, we need to solve for individual ranges for the given initial velocities. To do this, we shall use the formula

\text{R}=\frac{\text{v}_{\text{0}}^2\:\sin 2\theta _{\text{0}}}{\text{g}}

When the initial velocity is 30 m/s, the range is

\text{R}=\frac{\left(30\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=91.74\:\text{m}

When the initial velocity is 40 m/s, the range is

\text{R}=\frac{\left(40\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=163.10\:\text{m}

When the initial velocity is 50 m/s, the range is

\text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=254.84\:\text{m}

Based on the results, we can say that the ranges are approximately equal. The differences are only due to round-off errors.

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College Physics by Openstax Chapter 3 Problem 30

A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

Solution:

To illustrate the problem, consider the following figure:

A player passes the ball 7 meters across the field with an initial velocity of 12 m/s

Part A

We are given the 7-meter range, R, and the initial velocity, vo, of the projectile. We have R=7.0 m, and vo=12.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range

\text{R}=\frac{\text{v}^{2}_{\text{o}}\sin 2\theta _{\text{o}}}{g}

Solving for θo in terms of the other variables, we have

\begin{align*}
\text{gR} & =\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}} \\
\sin 2\theta _{\text{o}} & =\frac{\text{gR}}{\text{v}_{\text{o}}^2} \\
2\theta _\text{o} & =\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\
\theta _\text{o} & =\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\
\end{align*}

Substituting the given values, we have

\begin{align*}
\theta _\text{o} & =\frac{1}{2} \sin ^{-1}\left[\frac{\left(9.81\text{m/s}^2\right)\left(7.0\text{m}\right)}{\left(12.0\text{m/s}\right)^2}\right] \\

\theta _\text{o} & =14.2^{\circ}

\qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

Part B

The other angle that would give the same range is actually the complement of the solved angle in Part A. The other angle,

\theta _o'=90^{\circ} -14.24^{\circ} =75.8^{\circ} \qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\

This angle is not used as often, because the time of flight will be longer. In rugby that means the defense would have a greater time to get into position to knock down or intercept the pass that has the larger angle of release.

Part C

We can use the x-component of the motion to solve for the time of flight.

\Delta \text{x}=\text{v}_\text{x}\text{t}

We need the horizontal component of the velocity. We should be able to solve for the component since we are already given the initial velocity and the angle.

\text{v}_{\text{x}}=\left(12\:\text{m/s}\right)\cos 14.24^{\circ} =11.63\:\text{m/s}

Therefore, the total time of flight is

\begin{align*}
\text{t} & =\frac{\Delta \text{x}}{\text{v}_{\text{x}}} \\
\text{t} & =\frac{7.0\:\text{m}}{11.63\:\text{m/s}} \\
\text{t} & =0.60\:\text{s}

\qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}
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College Physics by Openstax Chapter 3 Problem 29

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

Solution:

To illustrate the problem, consider the following figure:

The archer and the target at 75 meter range

Part A

We are given the range of 75-meter range, R, and the initial velocity, vo, of the projectile. We have R=75.0 m, and vo=35.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range

\text{R}=\frac{\text{v}^{2}_{\text{o}}\:\sin 2\theta _{\text{o}}}{g}

Solving for θo in terms of the other variables, we have

\begin{align*}

\text{gR} & =\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}} \\
\sin \:2\theta _{\text{o}} & =\frac{\text{gR}}{\text{v}_{\text{o}}^2} \\
2\theta _\text{o} & =\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\
\theta _\text{o} & =\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\
\theta _o & =\frac{1}{2}\sin ^{-1}\left[\frac{\left(9.81\:\text{m/s}^2\right)\left(75.0\:\text{m}\right)}{\left(35.0\:\text{m/s}\right)^2}\right] \\
\theta _o & =18.46^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
 
\end{align*}

Part B

We know that halfway, the maximum height of the projectile occurs. Also at this instant, the vertical velocity is zero. We can solve for the maximum height and compare it with the given height of 3.50 meters.

The maximum height can be computed using the formula

\text{h}_{\text{max}}=\frac{\text{v}_{\text{oy}}^2}{2\text{g}}

To compute for the maximum height, we need the initial vertical velocity, voy. Since we know the magnitude and direction of the initial velocity, we have

\begin{align*}

\text{v}_{\text{oy}} & =\left(35.0\:\text{m/s}\right)\sin 18.46^{\circ} \\
\text{v}_{\text{oy}} & =11.08\:\text{m/s}
 
\end{align*}

Therefore, the maximum height is

\begin{align*}

\text{h}_{\max } & =\frac{\left(11.08\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)} \\
\text{h}_{\max } & =6.26\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

 
\end{align*}

We have known that the path of the arrow is above the branch of the tree. Therefore, the arrow will go through.

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College Physics by Openstax Chapter 3 Problem 28

(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed of 40.0 m/s (144 km/h) . How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)

Solution:

To illustrate the problem, consider the following figure:

The projectile path of the daredevil from the ramp

Part A

To determine the number of buses that the daredevil can clear, we will divide the range of the projectile path by 20 m, the length of 1 bus. That is

\text{no. of bus}=\frac{\text{Range}}{\text{bus length}}

First, we need to solve for the range.

\begin{align*}
\text{Range} & =\frac{\text{v}_{\text{o}}^2\:\sin 2\theta }{\text{g}} \\
\text{Range} & =\frac{\left(40.0\:\text{m/s}\right)^2\sin \left[2\left(32^{\circ} \right)\right]}{9.81\:\text{m/s}^2} \\
\text{Range} & =146.7\:\text{m} \\

\end{align*}

Therefore, the number of buses cleared is

\begin{align*}
\text{no. of buses} & =\frac{146.7\:\text{m}}{20\:\text{m}} \\
\text{no. of buses} & =7.34\:\text{buses} \\
\text{no. of buses} & =7\:\text{buses}

\qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

Therefore, he can only clear 7 buses. 

Part B

He clears the last bus by 6.7 m, which seems to be a large margin of error, but since we neglected air resistance, it really isn’t that much room for error.

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