Home Page Textbooks Physics: Principles and Problems Page 55: Standardized Test Practice

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

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Page 55: Standardized Test Practice

Exercise 1

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C- The dots would be close together to start with, and get farther apart as the plane accelerated. As the plane starts to move, it starts off slow, then begins to pick up speed (acceleration) in order to take lift. So in the particle model motion diagram, the gaps are close at first, then starts to spread out more.

Result

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The answer is C.

Exercise 2

Solution 1

Solution 2

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A vector diagram is need to solve all Physics problems

Result

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In this problem, we are asked to analyze a number of statements and determine which are false.

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**Statement A**

**A** states that to solve any physics problem we need to draw a vector diagram first.

This is incorrect, there are many problems concerning only vector intensities and scalar quantities where vector diagrams are not needed.

So statement **A** is false.

Step 3

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**Statement B**

**B** states that the length of a vector should be proportional to the data.

The drawn length of a vector must adequately represent the vector’s intensity. To accomplish this, the data describing the vector must be considered.

So statement **B** is true.

Step 4

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**Statement C**

**C** states that vectors can be added by simply adding their intensities(their lengths).

This is very incorrect, the fact that vectors cannot be added by just adding their intensities is, in a sense, what makes them vectors and not scalars.

Their direction has to be considered, and figures in the calculation.

So statement **C** is false.

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**Statement D**

**D** states that vectors can be added in either triangles or straight lines.

The statement is quite simply true.

When adding vectors, they can be drawn on the same straight line if they are parallel(or antiparallel) but must be drawn in a triangle if they point in different directions.

So statement **D** is true.

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**Conclusion**

Statements **A** and **C** are false.

Statements **B** and **D** are true.

Exercise 3

Step 1

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In a position vs. time diagram, velocity is the slope of the graph at a given time t.

Step 2

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Since the slope is the STEEPEST in the interval III, the answer is

Result

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Exercise 4

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In a position vs. time diagram, velocity is the slope of the graph at a given time t.

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Since the slope is positive all the way to point C, this means that the direction of motion was positive this whole time.

The cyclist was increasing his displacement (moving away from initial position) up to then.

At C, the velocity becomes negative, meaning that the cyclist turns around,

and in doing so, decreases his displacement as he heads back to initial position.

Answer: C

Step 3

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A simpler conclusion could be reached by observing that the graph gives the greatest value for C in terms of displacement.

So it is the farthest from inital point.

Result

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Exercise 5

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The distance traveled is the change in position in each interval.

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From least to greatest the change in position (distance traveled):
II, IV, III, I

(Section III and section I are very close, but section I seems to have a larger change in position on the graph)

Result

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Exercise 6

Step 1

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Graph A, because it is the only graph

Step 2

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that has initial vertical displacement equal to the final displacement.

(the squirrel returns to initial height)

Result

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Exercise 7

Step 1

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Observing vertical (north-south) and horizontal (east-west) components separately,

the rat displaces first 1.0m to the north, then later, 0.8m to the south.

Vertical component: 0.2m north.

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Horizontally: first displaces 0.3m east, then later, again east for 0.4m.

Total horizontal displacement: 0.7 m east.

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The final position in regard to the inital can be written as:

0.7m east, 0.2m north.

By placing the origin of a coordinate system at the initial point, (+x) being East, (+y) north, this position would represent the point with coordinates (0.7,0.2)m.

Its distance from the origin is calculated with the Pythagorean theorem:

$$
d=sqrt{0.7^2+0.2^2}=0.73m=73cm
$$

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