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Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 47: Section Review

Exercise 29

Step 1

1 of 2

To find slope, we would calculate rise over run; therefore, Slope of A = -2; Slope of B = $dfrac{3}{2}$; Slope of C = -1; Slope of D = 1

Step 2

2 of 2

For speed, you would use the absolute value of the slope. Therefore, the ranking from greatest average speed to least average speed would be A, B, C=D.

Exercise 30

Solution 1

Solution 2

Step 1

1 of 1

The average velocity will be the largest for the line with the largest slope (with sign)

$$
B>D>C>A
$$

Step 1

1 of 5

In this problem, we are asked to look at the given graph and compare the average velocities of all the objects.

Step 2

2 of 5

To find the average velocity of an object from the graph we need to divide its displacement by the time interval needed for that movement.

We will take into account the negative sign for those objects moving in the negative direction.

Step 3

3 of 5

We find the following velocities:
$$
begin{align*}
V_{text{A}} &= frac{d_{text{f}} – d_{text{i}}}{t_{text{f}} – t_{text{i}}}\
&= frac{(-3text{ m}) – 3text{ m}}{3text{ s} – 0text{ s}}\
&= – frac{6text{ m}}{3text{ s}}\
&= boxed{- 2 , frac{text{m}}{text{s}}}
end{align*}
$$
$$
begin{align*}
V_{text{B}} &= frac{d_{text{f}} – d_{text{i}}}{t_{text{f}} – t_{text{i}}}\
&= frac{3text{ m} – 0text{ m}}{2text{ s} – 0text{ s}}\
&= frac{3text{ m}}{2text{ s}}\
&= boxed{1.5 , frac{text{m}}{text{s}}}
end{align*}
$$
$$
begin{align*}
V_{text{C}} &= frac{d_{text{f}} – d_{text{i}}}{t_{text{f}} – t_{text{i}}}\
&= frac{(-3text{ m}) – 2text{ m}}{5text{ s} – 0text{ s}}\
&= – frac{5text{ m}}{5text{ s}}\
&= boxed{- 1 , frac{text{m}}{text{s}}}
end{align*}
$$
$$
begin{align*}
V_{text{D}} &= frac{d_{text{f}} – d_{text{i}}}{t_{text{f}} – t_{text{i}}}\
&= frac{4text{ m} – (-1text{ m})}{5text{ s} – 0text{ s}}\
&= frac{5text{ m}}{5text{ s}}\
&= boxed{1 , frac{text{m}}{text{s}}}
end{align*}
$$

Step 4

4 of 5

Now ranking the velocities from greatest to least we write:
$$
begin{gather*}
v_{text{B}} > v_{text{D}} > v_{text{C}} > v_{text{A}}\\
(1.5 > 1 > -1 > -2)
end{gather*}
$$

Result

5 of 5

$$text{B}>text{D}>text{C}>text{A}$$

Exercise 31

Step 1

1 of 4

$$
textbf{underline{textit{Solution}}}
$$

Step 2

2 of 4

If we examine the given graph we could notice that there is a 2 axes, and where usually the axes intersect at the origins, so will find the initial position with the respect to the origin “intersection of the two axes”.

Moreover, we don’t know the scale of the graph is $y$-axis, so instead of stating specifically the initial position of each object numerically we could state the initial position in terms of the graph is unit of scale, i.e. “1 square is a 1 unit of length”.

Also, we will take the upper part of the $y$-axis to be positive, i.e. the value of the position is positive, and the lower part is negative.

Thus, The initial position of each object is as follows:

* A: The initial position of this object is at a +ve 3 units of length “3 squares”.
* B: The initial position of this object is at the origin.
* C: The initial position of this object is at a +ve 2 units of length.
* D: The initial position of this object is a a -ve 1 units of length.

Thus, the arrangement of the initial position of the object from the most positive to the most negative is as follows :

A $>$ C $>$ B $>$ D

Step 3

3 of 4

If we had asked to arrange by the initial distance from the origin, we would have a different arrangement.

As the distance is a measure for how far is a point from the origin, it doesn’t matter whether it is positive or negative, i.e. we could say that the distance in our case is the absolute value of the initial position.

While the position is a location of something relative to the origin, so a 2 equidistant objects and a negative directions “one in the upper part and the other in the lower part of the $y$-axis”would have the same distance but not the same position.

Thus, the initial distance for each object is as follows:

* A: the initial distance from origin is 3 units of length.
* B: The initial distance from origin is zero.
* C: The initial distance from origin is 2 units of length.
* D: The initial distance from origin is 1 units of length.

A $>$ C $>$ D $>$ B

Result

4 of 4

* ize
* Arranging by initial position: A $>$ C $>$ B $>$ D
* Arranging by initial distance from origin: A $>$ C $>$ D $>$ B

Exercise 32

Solution 1

Solution 2

Step 1

1 of 2

absolute value of average velocity = average speed

Result

2 of 2

Average speed is the absolute value of average velocity. Average speed is only concerned with the rate of movement of an object, while average velocity is concerned with both the rate of movement and the direction in which the object is moving.

Step 1

1 of 4

In this problem, we are asked to explain the difference(the relation) between average velocity and average speed.

Step 2

2 of 4

Average velocity is a physical quantity that describes how a given object is moving.

It describes both the direction of the movement and the rate of displacement through time.

It is a **vector** quantity.

Step 3

3 of 4

Average speed is a physical quantity that describes only the rate of movement of a given object.

It is a **scalar** quantity.

Step 4

4 of 4

**Conclusion**

As we have explained, the average speed is the intensity of the average velocity vector.

Exercise 33

Solution 1

Solution 2

Step 1

1 of 2

Creating pictorial and physical models helps you get the situation organized and makes you visualize the proper units on the quantities you will substitute into the equation later on.

Result

2 of 2

Keeps the problem organized and helps visualize the proper units.

Step 1

1 of 3

In this problem, we are asked to explain why pictures and diagrams are important parts of solving any physical problem.

Step 2

2 of 3

Physics describes a wide variety of phenomena and behaviors using different equations and laws.

But some of those equations can be very abstract and mathematically cumbersome.

Step 3

3 of 3

In many cases, pictorial models and physical models(even rough ones) can make the situation more intuitive and clear.

Once the essence of the phenomenon is understood, the heavy mathematical artillery can be brought in for a detailed analysis.

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