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Schaum’s Outline of College Physics

Eugene Hecht

Chapter 1

Speed, Displacement, and Velocity: An Introduction to Vectors - all with Video Answers

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Chapter Questions

01:13

Problem 1

A toy train moves along a winding track at an average speed of $0.25$ $\mathrm{m} / \mathrm{s}$. How far will it travel in $4.00$ minutes? (See Appendix A on significant figures.)
The defining equation is $u_{a v}=l / t$. Here $l$ is in meters, and $t$ is in seconds, so the first thing to do is convert $4.00$ min into seconds:
$(4.00 \mathrm{~min})(60.0 \mathrm{~s} / \mathrm{min})=240 \mathrm{~s}$. Solving the equation for $l$
$$
l=v_{\text {av }} t=(0.25 \mathrm{~m} / \mathrm{s})(240 \mathrm{~s})
$$
Since the speed has only two significant figures, $l=60 \mathrm{~m}$.

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:06

Problem 2

A student driving a car travels $10.0 \mathrm{~km}$ in $30.0$ min. What was her average speed?

The defining equation is $u_{a v}=l / t$. Here $l$ is in kilometers, and $t$ is in minutes, so the first thing to do is convert $10.0 \mathrm{~km}$ to meters and then $30.0$ min into seconds: $(10.0 \mathrm{~km})(1000 \mathrm{~m} / \mathrm{km})=10.0 \times$
$10^{3} \mathrm{~m}$ and $(30.0 \mathrm{~min}) \times(60.0 \mathrm{~s} / \mathrm{min})=1800 \mathrm{~s}$. We need to solve
for $v_{a u}$, giving the numerical answer to three significant figures:
$$
v_{a v}=\frac{l}{t}=\frac{10.0 \times 10^{3} \mathrm{~m}}{1800 \mathrm{~s}}=5.56 \mathrm{~m} / \mathrm{s}
$$

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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00:55

Problem 3

Rolling along across the machine shop at a constant speed of $4.25$ $\mathrm{m} / \mathrm{s}$, a robot covers a distance of $17.0 \mathrm{~m}$. How long does that journey take?
Since the speed is constant the defining equation is $v=l / t$. Multiply both sides of this expression by $t$ and then divide both by $u:$
$$
t=\frac{l}{v}=\frac{17.0 \mathrm{~m}}{4.25 \mathrm{~m} / \mathrm{s}}=4.00 \mathrm{~s}
$$

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:54

Problem 4

Change the speed $0.200 \mathrm{~cm} / \mathrm{s}$ to units of kilometers per year. Use 365 days per year.
$$
0.200 \frac{\mathrm{cm}}{\mathrm{s}}=\left(0.200 \frac{\mathrm{cm} \mathrm{r}}{\mathrm{s}}\right)\left(10^{-5} \frac{\mathrm{km}}{\mathrm{cm}}\right)\left(3600 \frac{\mathrm{x}}{\mathrm{K}}\right)\left(24 \frac{\mathrm{K}}{\overline{\mathrm{d}}}\right)\left(365 \frac{\mathrm{d}}{\mathrm{y}}\right)=63.1 \frac{\mathrm{km}}{\mathrm{y}}
$$

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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02:55

Problem 5

A car travels along a road and its odometer readings are plotted $1.5$ against time in Fig. $1-9 .$ Find the instantaneous speed of the car at points $A$ and $B$. What is the car's average speed?
Because the speed is given by the slope $\Delta l / \Delta t$ of the tangent line, we take a tangent to the curve at point $A$. The tangent line is the curve itself in this case, since it's a straight line. For the triangle shown near $A$, we have
$$
\frac{\Delta l}{\Delta t}=\frac{4.0 \mathrm{~m}}{8.0 \mathrm{~s}}=0.50 \mathrm{~m} / \mathrm{s}
$$
This is the speed at point $A$ and it's also the speed at point $B$ and at every other point on the straight-line graph. It follows that $u=0.50$ $\mathrm{m} / \mathrm{s}=u_{a v}$. When the speed is constant the distance versus time curve is a straight line.

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:16

Problem 6

A kid stands $6.00 \mathrm{~m}$ from the base of a flagpole which is $8.00 \mathrm{~m}$ tall. Determine the magnitude of the displacement of the brass eagle on top of the pole with respect to the youngster's feet.
The geometry corresponds to a 3-4-5 right triangle (i.e., $3 \times 2-4$ $\times 2-5 \times 2$ ). Thus, the hypotenuse, which is the 5 -side, must be $10.0 \mathrm{~m}$ long, and that's the magnitude of the displacement.

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:22

Problem 7

A runner makes one complete lap around a 200 -m track in a time of $25 \mathrm{~s}$. What were the runner's $(a)$ average speed and $(b)$ average velocity?
(a) From the definition,
$$
\text { Average speed }=\frac{\text { Distance traveled }}{\text { Time taken }}=\frac{200 \mathrm{~m}}{25 \mathrm{~s}}=8.0 \mathrm{~m} / \mathrm{s}
$$
(b) Because the run ended at the starting point, the displacement vector from starting point to end point has zero length. Since $\overrightarrow{\mathbf{v}}_{w v}=\overrightarrow{\mathbf{s}} / t$
$$
\left|\overrightarrow{\mathbf{v}}_{a v}\right|=\frac{0 \mathrm{~m}}{25 \mathrm{~s}}=0 \mathrm{~m} / \mathrm{s}
$$

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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03:02

Problem 8

Using the graphical method, find the resultant of the following two displacements: $2.0 \mathrm{~m}$ at $40^{\circ}$ and $4.0 \mathrm{~m}$ at $127^{\circ}$, the angles being taken relative to the $+x$ -axis, as is customary. Give your answer to two significant figures. (See Appendix A on significant figures.)
Choose $x$ - and $y$ -axes as seen in Fig. $1-10$ and lay out the displacements to scale, tip to tail from the origin. Notice that all angles are measured from the $+x$ -axis. The resultant vector $\overrightarrow{\mathbf{s}}$ points from starting point to end point as shown. We measure its length on the scale diagram to find its magnitude, $4.6 \mathrm{~m}$. Using a protractor, we measure its angle $\theta$ to be $101^{\circ} .$ The resultant displacement is therefore $4.6 \mathrm{~m}$ at $101^{\circ}$.

Paul Gabriel
Paul Gabriel
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01:18

Problem 9

Find the $x$ - and $y$ -components of a 25.0-m displacement at an angle of $210.0^{\circ}$
The vector displacement and its components are depicted in Fig. 1-11. The scalar components are
$$
\begin{array}{l}
x \text { -component }=-(25.0 \mathrm{~m}) \cos 30.0^{\circ}=-21.7 \mathrm{~m} \\
y \text { -component }=-(25.0 \mathrm{~m}) \sin 30.0^{\circ}=-12.5 \mathrm{~m}
\end{array}
$$
Notice in particular that each component points in the negative coordinate direction and must therefore be taken as negative.

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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03:03

Problem 10

Solve Problem $1.8$ by use of rectangular components.
We resolve each vector into rectangular components as illustrated in Fig. $1-12(a)$ and $(b)$. (Place a cross-hatch symbol on the original vector to show that it is replaced by its components.) The resultant has scalar components of
$$
s_{x}=1.53 \mathrm{~m}-2.41 \mathrm{~m}=-0.88 \mathrm{~m} \quad s_{y}=1.29 \mathrm{~m}+3.19 \mathrm{~m}=4.48 \mathrm{~m}
$$
Notice that components pointing in the negative direction must be assigned a negative value. Thus, since $s x$ is to the left in the negative $x$ -direction it is negative, whereas sy is upward in the positive $y$ -direction and is positive.
The resultant is shown in $\underline{\text { Fig. } 1-12(c) \text { ; there, }}$
$$
s=\sqrt{(0.88 \mathrm{~m})^{2}+(4.48 \mathrm{~m})^{2}}=4.6 \mathrm{~m} \quad \tan \phi=\frac{4.48 \mathrm{~m}}{0.88 \mathrm{~m}}
$$
and $\varphi=79^{\circ}$, from which $\theta=180^{\circ}-\varphi=101^{\circ} .$ Hence, $\overrightarrow{\mathrm{s}}=4.6 \mathrm{~m}$ -
$101^{\circ}$ FROM $+x$ -AXIS; remember, vectors must have their directions stated explicitly.

Paul Gabriel
Paul Gabriel
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02:28

Problem 11

Add the following two displacement vectors using the parallelogram method: $30 \mathrm{~m}$ at $30^{\circ}$ and $20 \mathrm{~m}$ at $140^{\circ} .$ Remember that numbers like $30 \mathrm{~m}$ and $20 \mathrm{~m}$ have two significant figures.
The vectors are drawn with a common origin in Fig. $1-13(a) .$ We construct a parallelogram using them as sides, as shown in $\underline{\text { Fig. } 1-}$ $13(b)$. The resultant $\vec{s}$ is then represented by the diagonal. By measurement, we find that $\overrightarrow{\mathrm{s}}$ is $30 \mathrm{~m}$ at $69^{\circ}$.

Paul Gabriel
Paul Gabriel
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02:12

Problem 12

Express the vectors illustrated in Figs. $1-12(c), \underline{1-14}, \underline{1-15}$, and $\underline{1-16}$ in the form $\overrightarrow{\mathbf{R}}=R_{x} \hat{\mathbf{i}}+R_{y} \hat{\mathbf{j}}+R_{z} \hat{\mathbf{k}}$ (leave out the units). If you are
not using basis vectors skip this problem. Remembering that plus and minus signs must be used to show direction along an axis,
$$
\begin{aligned}
\text { For Fig. } 1-12(c): & & \overrightarrow{\mathbf{R}} &=-0.88 \hat{\mathbf{i}}+4.48 \hat{\mathbf{j}} \\
\text { For Fig. 1-14: } & & \overrightarrow{\mathbf{R}}=5.7 \hat{\mathbf{i}}-3.2 \hat{\mathbf{j}} \\
\text { For Fig. 1-15: } & & \overrightarrow{\mathbf{R}}=-94 \hat{\mathbf{i}}+71 \hat{\mathbf{j}} \\
\text { For Fig. 1-16: } & & \overrightarrow{\mathbf{R}}=46 \hat{\mathbf{i}}+39 \hat{\mathbf{j}}
\end{aligned}
$$

Paul Gabriel
Paul Gabriel
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03:14

Problem 13

Perform graphically the following vector additions and subtractions, where $\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{B}}$, and $\overrightarrow{\mathbf{C}}$ are the vectors drawn in $\underline{\text { Fig. } 1-}$ $\underline{17}:(a) \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} ;(b) \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}} ;(c) \overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}} ;(d) \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{C}} .$
See $\underline{\text { Fig. } 1-16}(a)$ through $(d) .$ In $(c), \overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{A}}+(-\overrightarrow{\mathbf{B}})$; that is, to
subtract $\overrightarrow{\mathbf{B}}$ from $\overrightarrow{\mathbf{A}}$, reverse the direction of $\overrightarrow{\mathbf{B}}$ and add it vectorially to $\overrightarrow{\mathbf{A}} .$ Similarly, in $(d), \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{C}}=\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+(-\overrightarrow{\mathbf{C}})$, where $-\overrightarrow{\mathbf{C}}$ is
equal in magnitude but opposite in direction to $\overrightarrow{\mathbf{C}}$.

Paul Gabriel
Paul Gabriel
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01:59

Problem 14

$\overrightarrow{\mathbf{B}}$. If you have not learned to use basis vectors, skip this problem.
From a purely mathematical approach,
$$
\begin{aligned}
\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}} &=(-3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})-(-12 \hat{\mathbf{i}}+25 \hat{\mathbf{j}}+13 \hat{\mathbf{k}}) \\
&=-3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}+12 \hat{\mathbf{i}}-25 \hat{\mathbf{j}}-13 \hat{\mathbf{k}}=12 \hat{\mathbf{i}}-28 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}
\end{aligned}
$$
Notice that $_{1 j-2 j-1 z}$ is simply $\overrightarrow{\mathbf{A}}$ reversed in direction. Therefore, we have, in essence, reversed $\overrightarrow{\mathbf{A}}$ and added it to $\overrightarrow{\mathbf{B}}$.

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:26

Problem 15

A boat can travel at a speed of $8 \mathrm{~km} / \mathrm{h}$ in still water on a lake. In the flowing water of a stream, it can move at $8 \mathrm{~km} / \mathrm{h}$ relative to the water in the stream. If the stream speed is $3 \mathrm{~km} / \mathrm{h}$, how fast can the boat move past a tree on the shore when it is traveling $(a)$ upstream and $(b)$ downstream?
(a) If the water was standing still, the boat's speed past the tree would be $8 \mathrm{~km} / \mathrm{h}$. But the stream is carrying it in the opposite direction at $3 \mathrm{~km} / \mathrm{h}$. Therefore, the boat's speed relative to the tree is $8 \mathrm{~km} / \mathrm{h}-3 \mathrm{~km} / \mathrm{h}=5 \mathrm{~km} / \mathrm{h}$.
(b) In this case, the stream is carrying the boat in the same direction the boat is trying to move. Hence, its speed past the tree is $8 \mathrm{~km} / \mathrm{h}+3 \mathrm{~km} / \mathrm{h}=11 \mathrm{~km} / \mathrm{h}$.

Paul Gabriel
Paul Gabriel
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01:56

Problem 16

A plane is traveling eastward at an airspeed of $500 \mathrm{~km} / \mathrm{h}$. But a 90$\mathrm{km} / \mathrm{h}$ wind is blowing southward. What are the direction and speed of the plane relative to the ground? If you have not learned how to deal with relative velocities, skip this problem.

The plane's resultant velocity with respect to the ground, ${ }_{\text {PG }}$, is the sum of two vectors, the velocity of the plane with respect to the air, $_{\text {iPA }}=500 \mathrm{~km} / \mathrm{h}$ - EAST and the velocity of the air with respect to the ground, ${ }_{* \mathrm{AG}}=90 \mathrm{~km} / \mathrm{h}$ - SOUTH. In other words, ${ }_{\mathrm{PG}}={ }_{\text {'PA }}$ AG. These component velocities are shown in Fig. 1-18. The plane's resultant speed is
$$
v_{\mathrm{PG}}=\sqrt{(500 \mathrm{~km} / \mathrm{h})^{2}+(90 \mathrm{~km} / \mathrm{h})^{2}}=508 \mathrm{~km} / \mathrm{h}
$$
The angle $\alpha$ is given by
$$
\tan \alpha=\frac{90 \mathrm{~km} / \mathrm{h}}{500 \mathrm{~km} / \mathrm{h}}=0.18
$$
from which $\alpha=10^{\circ}$. The plane's velocity relative to the ground is $508 \mathrm{~km} / \mathrm{h}$ at $10^{\circ}$ south of east.

Paul Gabriel
Paul Gabriel
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01:20

Problem 17

With the same airspeed as in Problem $1.16$, in what direction must the plane head in order to move due east relative to the Earth?
The sum of the plane's velocity through the air and the velocity of the wind will be the resultant velocity of the plane relative to the Earth. This is shown in the vector diagram in Fig. $1-19 .$ Notice that, as required, the resultant velocity is eastward. Keeping in mind that the wind speed is given to two significant figures, it is seen that $\sin \theta=(90 \mathrm{~km} / \mathrm{h})(500 \mathrm{~km} / \mathrm{h})$, from which $\theta=10^{\circ}$. The plane should head $10^{\circ}$ north of east if it is to move eastward relative to the Earth.
To find the plane's eastward speed, we note in the figure that $U_{\mathrm{PG}}$ $=(500 \mathrm{~km} / \mathrm{h}) \cos \theta=4.9 \times 10^{5} \mathrm{~m} / \mathrm{h}$

Paul Gabriel
Paul Gabriel
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00:34

Problem 18

Three kids in a parking lot launch a rocket that rises into the air along a $380-\mathrm{m}$ long arc in $40 \mathrm{~s}$. Determine its average speed.

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:18

Problem 19

An ant walked $10.0 \mathrm{~cm}$ across the floor in $6.2 \mathrm{~s}$. What was its average speed in $\mathrm{m} / \mathrm{s}$ ? [Hint: 2 significant figures. You are given $s$ and $t$ and must find uau. Watch out for units.]

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:24

Problem 20

A 12-mg housefly has a maximum speed of $4.5$ mph; what is that in $\mathrm{m} / \mathrm{s}$ ? [ Hint: 2 significant figures. $1 \mathrm{mph}=0.44707 \mathrm{~m} / \mathrm{s}$.]
$\mathbf{1 . 2 1}$ [I] According to its computer, a robot that left its closet and traveled $1200 \mathrm{~m}$, had an average speed of $20.0 \mathrm{~m} / \mathrm{s}$. How long did the trip take?

Paul Gabriel
Paul Gabriel
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02:03

Problem 22

A car's odometer reads $22687 \mathrm{~km}$ at the start of a trip and 22791 $\mathrm{km}$ at the end. The trip took $4.0$ hours. What was the car's average speed in $\mathrm{km} / \mathrm{h}$ and in $\mathrm{m} / \mathrm{s}$ ?

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:11

Problem 23

A model plane flew $100 \mathrm{~m}$ in $25.0 \mathrm{~s}$ followed by another $240 \mathrm{~m}$ in an additional $60.0 \mathrm{~s}$, whereupon it crashed into the ground. How far did it travel in total? How long was it in the air? What was its average speed? [Hint: The overall average is not equal to the average of the averages. When you have several segments in a problem, label them like this: $l_{1}$ and $l_{2}$ and $t_{1}$ and $t_{2}$, such that $l=l_{1}$ $+l_{2}$ and $\left.t=t_{1}+t_{2} .\right]$

Paul Gabriel
Paul Gabriel
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01:45

Problem 24

A toy car traveled at an average speed of $2.0 \mathrm{~m} / \mathrm{s}$ for $20 \mathrm{~s}$, followed by $40 \mathrm{~s}$ at an average speed of $1.0 \mathrm{~m} / \mathrm{s}$, whereupon it came to a stop. How far in total did it go? How long in time did it travel? What was its average speed?

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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03:15

Problem 25

An auto travels at the rate of $25 \mathrm{~km} / \mathrm{h}$ for $4.0$ minutes, then at 50 $\mathrm{km} / \mathrm{h}$ for $8.0$ minutes, and finally at $20 \mathrm{~km} / \mathrm{h}$ for $2.0$ minutes. Find
(a) the total distance covered in $\mathrm{km}$ and $(b)$ the average speed for the complete trip in $\mathrm{m} / \mathrm{s}$.

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:22

Problem 26

Starting from the center of town, a car travels east for $80.0 \mathrm{~km}$ and then turns due south for another $192 \mathrm{~km}$, at which point it runs out of gas. Determine the displacement of the stopped car from the center of town.

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:25

Problem 27

A little turtle is placed at the origin of an $x y$ -grid drawn on a large sheet of paper. Each grid box is $1.0 \mathrm{~cm}$ by $1.0 \mathrm{~cm}$. The turtle walks around for a while and finally ends up at point $(24,10)$, that is, 24 boxes along the $x$ -axis, and 10 boxes along the $y$ -axis. Determine the displacement of the turtle from the origin at the point.

Paul Gabriel
Paul Gabriel
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03:38

Problem 28

A bug starts at point $A$, crawls $8.0 \mathrm{~cm}$ east, then $5.0 \mathrm{~cm}$ south, $3.0$
$\mathrm{cm}$ west, and $4.0 \mathrm{~cm}$ north to point $B$. $(a)$ How far south and east is $B$ from $A$ ? $(b)$ Find the displacement from $A$ to $B$ both graphically and algebraically.

Paul Gabriel
Paul Gabriel
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01:56

Problem 29

A runner travels $1.5$ laps around a circular track in a time of $50 \mathrm{~s}$. The diameter of the track is $40 \mathrm{~m}$ and its circumference is $126 \mathrm{~m}$. Find $(a)$ the average speed of the runner and $(b)$ the magnitude of the runner's average velocity. Be careful here; average speed depends on the total distance traveled, whereas average velocity depends on the displacement at the end of the particular journey.

Paul Gabriel
Paul Gabriel
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01:18

Problem 30

During a race on an oval track, a car travels at an average speed of $200 \mathrm{~km} / \mathrm{h}$. (a) How far did it travel in $45.0$ min? (b) Determine its average velocity at the end of its third lap.

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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06:23

Problem 31

The following data describe the position of an object along the $x$ axis as a function of time. Plot the data, and find, as best you can, the instantaneous velocity of the object at $(a) t=5.0 \mathrm{~s},(b) 16 \mathrm{~s}$, and $(c) 23 \mathrm{~s} .$

Brandy Heflin
Brandy Heflin
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03:45

Problem 32

For the object whose motion is described in Problem $1.27$, as best you can, find its velocity at the following times: (a) $3.0 \mathrm{~s}$, (b) $10 \mathrm{~s}$, and (c) $24 \mathrm{~s}$.

Morgan Cheatham
Morgan Cheatham
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01:58

Problem 33

Find the scalar $x$ - and $y$ -components of the following displacements in the $x y$ -plane: (a) $300 \mathrm{~cm}$ at $127^{\circ}$ and
(b) $500 \mathrm{~cm}$ at $220^{\circ}$.

Paul Gabriel
Paul Gabriel
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04:17

Problem 34

Starting at the origin of coordinates, the following displacements are made in the $x y$ -plane (that is, the displacements are coplanar):
$60 \mathrm{~mm}$ in the $+y$ -direction, $30 \mathrm{~mm}$ in the $-x$ -direction, $40 \mathrm{~mm}$ at $150^{\circ}$, and $50 \mathrm{~mm}$ at $240^{\circ}$. Find the resultant displacement both graphically and algebraically.

Paul Gabriel
Paul Gabriel
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04:36

Problem 35

Compute algebraically the resultant of the following coplanar displacements: $20.0 \mathrm{~m}$ at $30.0^{\circ}, 40.0 \mathrm{~m}$ at $120.0^{\circ}, 25.0 \mathrm{~m}$ at $180.0^{\circ}, 42.0 \mathrm{~m}$ at $270.0^{\circ}$, and $12.0 \mathrm{~m}$ at $315.0^{\circ}$. Check your answer with a graphical solution.

Paul Gabriel
Paul Gabriel
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01:36

Problem 36

What displacement at $70^{\circ}$ has an $x$ -component of $450 \mathrm{~m}$ ? What is its $y$ -component?

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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04:48

Problem 37

What displacement must be added to a 50 -cm displacement in the $+x$ -direction to give a resultant displacement of $85 \mathrm{~cm}$ at $25^{\circ}$ ?

Paul Gabriel
Paul Gabriel
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04:10

Problem 38

Refer to $\underline{\text { Fig. } 1-20 . \text { In terms of vectors } \overrightarrow{\mathbf{A}} \text { and } \overrightarrow{\mathbf{B}} \text { , express the }}$ vectors $(a)_{i},(b) \overrightarrow{\mathrm{k}},(c)_{i}$, and $(d)_{j} .$

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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04:24

Problem 39

Refer to Fig. 1-21. In terms of vectors $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$, express the vectors $(a)_{i},(b)_{i}-\overrightarrow{\mathbf{C}}$, and $(c)_{i}+_{i}-\overrightarrow{\mathbf{C}} .$

Emily Anderson
Emily Anderson
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02:13

Problem 40

$\operatorname{Find}(a) \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}},(b) \overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}$, and $(c) \overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{C}}$ if $\overrightarrow{\mathbf{A}}=7 \hat{\mathbf{i}}-6 \hat{\mathbf{j}} \cdot \overrightarrow{\mathbf{B}}=-3 \hat{\mathbf{i}}+12 \hat{\mathbf{j}} \boldsymbol{y}$ and $\overrightarrow{\mathbf{C}}=4 \hat{\mathbf{i}}-4 \hat{\mathbf{j}} \cdot$

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:04

Problem 41

Find the magnitude and angle of $\overrightarrow{\mathbf{R}}$ if $_{\mathrm{k}=1, \mathrm{in}-1 \mathrm{z}^{2}}$

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:13

Problem 42

Determine the displacement vector that must be added to the displacement $(25 \hat{\mathrm{i}}-16 \hat{j})$ $\mathrm{m}$ to give a displacement of $7.0 \mathrm{~m}$ pointing in the $+x$ -direction?

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:21

Problem 43

A vector $(1 \hat{s}-16 \hat{j}+27 \hat{k}$ is added to a vector $(23 \hat{\mathbf{j}}-40 \hat{\mathbf{k}}) \bullet$ What is the magnitude of the resultant?

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:53

Problem 44

A truck is moving north at a speed of $70 \mathrm{~km} / \mathrm{h}$. The exhaust pipe above the truck cab sends out a trail of smoke that makes an angle of $20^{\circ}$ east of south behind the truck. If the wind is blowing directly toward the east, what is the wind speed at that location? [Hint: The smoke reveals the direction of the truck with-respect-to the air.]

Supratim Pal
Supratim Pal
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01:22

Problem 45

A ship is traveling due east at $10 \mathrm{~km} / \mathrm{h}$. What must be the speed of a second ship heading $30^{\circ}$ east of north if it is always due north of the first ship?

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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02:33

Problem 46

A boat, propelled so as to travel with a speed of $0.50 \mathrm{~m} / \mathrm{s}$ in still water, moves directly across a river that is $60 \mathrm{~m}$ wide. The river flows with a speed of $0.30 \mathrm{~m} / \mathrm{s}$. (a) At what angle, relative to the straight-across direction, must the boat be pointed? (b) How long does it take the boat to cross the river?

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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01:27

Problem 47

A reckless drunk is playing with a gun in an airplane that is going directly east at $500 \mathrm{~km} / \mathrm{h}$. The drunk shoots the gun straight up at the ceiling of the plane. The bullet leaves the gun at a speed of $1000 \mathrm{~km} / \mathrm{h}$. According to someone standing on the Earth, what angle does the bullet make with the vertical?

Manish Kumar ( Iit K )
Manish Kumar ( Iit K )
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