homework assignment 4.1 judging space in seconds

Additional problems from Serway's fourth edition

(4 ed) 2.2 The new BMW M3 can accelerate from zero to 60 mi/h in 5.6 s

(a) What is the resulting acceleration in m/s 2 ?

(b) How long would it take the BMW to go from 60 mi/h to 130 mi/h at this rate?

(4 ed) 2.3 A hot air balloon is traveling vertically upward at a constant speed of 5.00 m/s. When it is 21.0 m above the ground, a package is released from the balloon.

(a) After it is released, for how long is the package in the air?

(b) What is its velocity just before impact with the ground!

(c) Repeat (a) and (b) for the case of the baloon descending at 5.0 m/s.

(4 ed) 2.4 A hockey player is standing on his skates on a frozen pond when an opposing player skates by with the puck, moving with a uniform speed of 12.0 m/s. After 3.00 s, the first player makes up his mind to chase his opponent. If the first player accelerates uniformly at 4.00 m/s 2 ,

(a) how long does it take him to catch the opponent?

(b) How far has the first player traveled in this time?

Conceptual Questions

2.Q4 Is it possible to have a situation in which the velocity and acceleration have opposite signs? If so, sketch a velocity-time graph to prove your point.

Certainly. Consider a car moving to the right but slowing down. Moving to the right means its velocity is positive. Slowing down means its velocity is de creasing or the change in velocity is negative and that means the acceleration is negative.

2.Q5 If the velocity of a particle is nonzero, can its acceleration be zero? Explain.

If the velocity is constant the acceleration is zero .

2.Q6 If the velocity of a particle is zero, can its acceleration be nonzero? Explain.

A velocity of zero is also a constant velocity and that means the acceleration is zero .

2.Q9 A student at the top of a building of height h throws one ball upward with an initial speed v yi and then throws a second ball downward wit the same initial speed. How do the final speeds of the balls compare when they reach the ground?

This one will be fun or interesting to talk about again after we have studied Energy Conservation.

When the ball that was initially thrown up ward comes back to the height of the top of the building, its speed is again v yi . It will have the same speed . Of course its velocity will be - v yi because it is moving down . That means that it has exactly the same speed and velocity at the ball that is initially thrown down ward with initial speed v yi so the two balls hit the ground with exactly the same speed (and velocity!).

2.Q16 A pebble is dropped into a water well and the splash is heard sixteen seconds later; as illustrated in the "BC" cartoon in Figure !2.16. Estimate the distance from the top of the well to the water's surface.

We can use the equation

y = y i + v yi + (1/2) a y t 2

We are measuring distances from the top of the well so y i = 0 and we drop the pebble so v yi = 0 so this equation reduces to

y = (1/2) a y t 2

y = (1/2) ( - 9.8 m/s 2 ) (16 s) 2

y = - 1 254.4 m

That is a very deep "well", indeed! But, maybe, wells were deeper in such prehistoric times. If we had used the approximation that a y = - g = - 10 m/s 2 , then our value would have been

y = - 1 280 m

Problems from the current (5th) edition of Serway and Beichner.

2.21 Jules Verne in 1865 proposed sending people to the Moon by firing a space capsule from a 220-m-long cannon with a final velocity of 10.97 km/s. What would have been the unrealistically large acceleration experienced by the space travelers during launch? Compare your answer with the free-fall acceleration of 9.8 m/s 2 .

First, let's change the final velocity to units of m/s; we can almost do that in our heads.

v f = 10.97 km / s [ 1000 m / km ] = 10.97 x 10 3 m / s

Solutions to the additional problems from Serway's fourth edition

a) t 1 ; the velocity, as the slope of the tangent line, is zero

b) t 2 ; the velocity, as the slope of the tangent line, is negative

c) t 3 ; the velocity, as the slope of the tangent line, is positive

d) t 4 ; the velocity, as the slope of the tangent line, is zero

a = v / t = [ 60 mi/h ] / 5.6 s = 10.7 (mi/h)/s a = 10.7 mi/h/s a = 10.7 mi /( h - s ) [ 1.61 km / mi ] [ 1000 m / km ] [ h / 3600 s ] = 4.79 m/s 2 a = 4.79 m/s 2

v = v i + a t a t = v - v i t = [v - v i ] / a t = [ 130 mi/h - 60 mi/h ] / (10.7 mi/h/s ) t = [ 70 mi/h ] / (10.7 mi/h/s ) t = 6.54 s

That is 6.45 seconds beyond the 5.6 s required to reach 60 mi/h. The total time will be

t tot = 5.6 s + 6.54 s t tot = 12.1 s

Once released, the package is in free fall with an acceleration of a = - g = - 9.8 m/s 2

We know its initial velocity and initial position

v i = 5 m/s y i = 21 m

The later position of the package is given by

y = y i + v i t + ( 1 / 2 ) a t 2 y = 21 m + (5 m/s) t + ( 1 / 2 ) ( - 9.8 m/s 2 ) t 2

Now we set y = 0 and solve for t

y = 0 = 21 m + (5 m/s) t + ( 1 / 2 ) ( - 9.8 m/s 2 ) t 2

As before, we can carry the units explicitly or we can ensure that we have consistent units and drop them and write only

4.9 t 2 - 5 t - 21 = 0

From the quadradic equation, we find the two solutions of t

t 1 = 2.64 s, and t 2 = - 1.62 s

Physically, we are only interested in solutions for t > 0. Mathematically, our equation is only valid for t > 0 since it is valid only after the package is released. So we use only t 1 .

Once released, the velocity of the package is given by v = v i + a t v = 5 m/s + ( - 9.8 m/s 2 ) t v = 5 m/s + ( - 9.8 m/s 2 ) (2.64 s) v = - 20.9 m/s

Of course, the minus sign indicates that the velocity is directed downward .

As before, once released, the package is in free fall with an acceleration of a = - g = - 9.8 m/s 2

v i = - 5 m/s

The package, along with the balloon, is now moving downward and this shows up as the negative sign on the velocity

y = y i + v i t + ( 1 / 2 ) a t 2 y = 21 m + ( - 5 m/s) t + ( 1 / 2 ) ( - 9.8 m/s 2 ) t 2

y = 0 = 21 m + ( - 5 m/s) t + ( 1 / 2 ) ( - 9.8 m/s 2 ) t 2

4.9 t 2 + 5 t - 21 = 0

t 1 = 1.62 s, and t 2 = - 2.64 s

Once released, the velocity of the package is given by

v = v i + a t v = - 5 m/s + ( - 9.8 m/s 2 ) t v = - 5 m/s + ( - 9.8 m/s 2 ) (1.62 s) v = - 20.9 m/s

Notice that the velocities are the same. Later on, we can describe this in terms of energy conservation. The kinetic energy of the package is the same whether it is thrown up with v = + 5 m/s or if it is thrown down with v = - 5 m/s.

We know the initial velocities and accelerations of the two players, a 1 = 4 m/s 2 , a 2 = 0 v 1i = 0, v 2i = 12 m/s x 1i = x 2i = 0

The position of player #2 is given by

x 2 = x 2i + v 2i t + ( 1 / 2 ) a 2 t 2 x 2 = (12 m/s) t

Be careful with the time . We must account for player #1's wait of 3 s. With this accounted for, we can calculate the position of player #1 from

x 1 = x 1i + v 1i (t - 3 s) + ( 1 / 2 ) a 1 (t - 3 s) 2

Of course, this equation only makes sense for t > 3 s.

x 1 = 0 + 0 + ( 1 / 2 ) (4 m/s 2 ) (t - 3 s) 2 = (2 m/s 2 ) (t 2 - 6 s t + 9 s 2 )

Now we set x 1 = x 2 and solve for the time t.

x 1 = x 2 (2 m/s 2 ) (t 2 - 6 s t + 9 s 2 ) = (12 m/s) t

We can either keep the units in explicitly or ensure that we have consistent units and simply write

2 (t 2 - 6t + 9) = 12t 2 t 2 - 12 t + 18 = 12 t 2 t 2 - 24 t + 18 = 0 t 2 - 12 t + 9 = 0

There are two solutions to this quadratic equation,

t 1 = 11.2 s, and t 2 = 0.8 s

However, the equation for the position of player #1 is not valid for t 2 < 3 s, so we keep only t x ,

Now where is player #2 (and, therefore, player #1 as well) at this time? x 2 = x 2i + v 2i t + ( 1 / 2 ) a 2 t 2 x 2 = (12 m/s) t x 2 = (12 m/s) (11.2 s) x 2 = 134.4 m

(Is a hockey rink that large?)

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4.1: Writing Assignment

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• Page ID 73568

• Precha Thavikulwat
• Towson University

The purpose of a case writing assignment is to give practice in the kind of writing that is valued in the workplace. If after you have competed your assignment, you would not yourself pay money for what you have written, then what you have written is not good enough.

In the workplace, you would not pay for writing that tells you facts you already know. The writer might mention what you already know here and there in the piece, but if the piece is substantially about facts you already know, the piece would have no value to you.

You would not pay for writing that is unclear, especially when you know the author could be clearer. If the author writes that the company should focus on something, you would want the author to clarify. How is the company supposed to focus? What action must be taken to constitute focus?

You would not pay for writing that tells you to research or analyze the problem. The assignment requires analyze for which research might be useful. Admonishing the reader to do research and analysis, however, is condescending and useless.

Generally, the assignment will ask that the case be written from the viewpoint of a consultant, an outsider of the firm. From this viewpoint, the firm is identified by name. Do not use first- and second-person pronouns, we and you, to refer to the firm.

Generally, the assignment will require a narrative. If so, write in complete sentences organized into paragraphs, with attention to topic sentences and transitions between paragraphs. Do not submit a list when a narrative is required. Whereas a narrative shows how the writer thinks, a list only shows that the writer knows some words.

The following is a checklist of additional pointers on case writing:

• Respond directly to the assignment. If the assignment is to submit research findings, do the research, submit the findings, and explain how you got the findings. If the assignment is to make a recommendation, analyze the facts, make a recommendation, and explain why the recommended course of action is best.
• Answer the question before explaining the answer. If the requirement is to answer in one paragraph, the answer should be the first sentence of the paragraph. If the requirement is to answer in a short essay, the answer should be the last sentence of the first paragraph. In this instance, the first one or two sentences of the first paragraph should lead to the answer, the middle paragraphs should explain the answer, and the last paragraph should re-iterate the answer.
• Cons before pros. Address reasonable arguments against your position before addressing the arguments for your position. The best argument for your position should be the last argument.
• Write about what you know. Write especially about what you know that others may not know, because they have not studied the case as well as you have studied it. The mind needs time to work, so study the case several days before you write about it.
• Do not question yourself or undermine your own position by apologizing, equivocating, or otherwise suggesting ineptitude. Do not write “In my opinion” (suggesting it is not worth much), “I think” (suggesting you do not know), or “I feel” (suggesting you do not think).
• Do not complain about insufficient information, or call for further study or research. The task of strategic management is to identify the best course of immediate action given available information.
• Write with grace and dignity. Do not disparage, patronize, preach, use slang, or make frivolous comments.
• Be specific. Business is a numbers game, so use numbers whenever possible. When describing a course of action, give enough detail for the action to be visualized, as in a skit. Action that cannot be visualized cannot be executed.
• Be assertive, neither wishy-washy nor dogmatic. Avoid mights (wishy-washy) and musts (dogmatic).
• Economize on words. Do not merely restate the facts of the case. Either use the facts, to support an argument or introduce a topic, or do not mention them at all. Do not write a long introduction—get quickly to the point.

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Chapter 6: Polynomials

6.3 Scientific Notation (Homework Assignment)

Scientific notation is a convenient notation system used to represent large and small numbers. Examples of these are the mass of the sun or the mass of an electron in kilograms. Simplifying basic operations such as multiplication and division with these numbers requires using exponential properties.

Scientific notation has two parts: a number between one and nine and a power of ten, by which that number is multiplied.

[latex]\text{Scientific notation: }a \times 10^b, \text{ where }1 \le a \le 9[/latex]

The exponent tells how many times to multiply by 10. Each multiple of 10 shifts the decimal point one place. To decide which direction to move the decimal (left or right), recall that positive exponents means there is big number (larger than ten) and negative exponents means there is a small number (less than one).

Example 6.3.1

Convert 14,200 to scientific notation.

[latex]\begin{array}{rl} 1.42&\text{Put a decimal after the first nonzero number} \\ \times 10^4 & \text{The exponent is how many times the decimal moved} \\ 1.42 \times 10^4& \text{Combine to yield the solution} \end{array}[/latex]

Example 6.3.2

Convert 0.0028 to scientific notation.

[latex]\begin{array}{rl} 2.8&\text{Put a decimal after the first nonzero number} \\ \times 10^{-3}&\text{The exponent is how many times the decimal moved} \\ 2.8\times 10^{-3}&\text{Combine to yield the solution} \end{array}[/latex]

Example 6.3.3

Convert 3.21 × 10 5 to standard notation.

Starting with 3.21, Shift the decimal 5 places to the right, or multiply 3.21 by 10 5 .

321,000 is the solution.

Example 6.3.4

Convert 7.4 × 10 −3 to standard notation

Shift the decimal 3 places to the left, or divide 6.4 by 10 3 .

0.0074 is the solution.

Working with scientific notation is easier than working with other exponential notation, since the base of the exponent is always 10. This means that the exponents can be treated separately from any other numbers. For instance:

Example 6.3.5

Multiply (2.1 × 10 −7 )(3.7 × 10 5 ).

First, multiply the numbers 2.1 and 3.7, which equals 7.77.

Second, use the product rule of exponents to simplify the expression 10 −7 × 10 5 , which yields 10 −2 .

Combine these terms to yield the solution 7.77 × 10 −2 .

Example 6.3.6

(4.96 × 10 4 ) ÷ (3.1 × 10 −3 )

First, divide: 4.96 ÷ 3.1 = 1.6

Second, subtract the exponents (it is a division): 10 4− −3 = 10 4 + 3 = 10 7

Combine these to yield the solution 1.6 × 10 7 .

For questions 1 to 6, write each number in scientific notation.

For questions 7 to 12, write each number in standard notation.

• 2.56 × 10 2

For questions 13 to 20, simplify each expression and write each answer in scientific notation.

• (7 × 10 −1 )(2 × 10 −3 )
• (2 × 10 −6 )(8.8 × 10 −5 )
• (5.26 × 10 −5 )(3.16 × 10 −2 )
• (5.1 × 10 6 )(9.84 × 10 −1 )
• [latex]\dfrac{(2.6 \times 10^{-2})(6 \times 10^{-2})}{(4.9 \times 10^1)(2.7 \times 10^{-3})}[/latex]
• [latex]\dfrac{(7.4 \times 10^4)(1.7 \times 10^{-4})}{(7.2 \times 10^{-1})(7.32 \times 10^{-1})}[/latex]
• [latex]\dfrac{(5.33 \times 10^{-6})(9.62 \times 10^{-2})}{(5.5 \times 10^{-5})^2}[/latex]
• [latex]\dfrac{(3.2 \times 10^{-3})(5.02 \times 10^0)}{(9.6 \times 10^3)^{-4}}[/latex]

Answer Key 6.3

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Lesson Three: Managing Space and Judging Time (4.3)

In previous chapters, you have learned how to search, identify, and target the space ahead of you. Knowing what is happening around you as you drive helps you make decisions about speed, lane position, and communication options.

Managing risk with SIM includes following distance . Maintaining an adequate following distance allows you to see restrictions ahead and respond in enough time to adjust your speed or lane position. Failing to maintain adequate following distance removes your control over the situation. You cannot respond; you can only react.

The minimum following distance is _______ seconds.

When at highway speeds, with vision restrictions , on slick roads, and behind motorcycles and/or large trucks, increase to _________ seconds.

Following Distance

Following too close creates problems for drivers. What are some problems that can be encountered when following too close?

  Point of No Return

Closure rate is the rate at which you are getting closer to the vehicle ahead, the space is getting smaller, decreasing. The vehicle ahead is not going the same speed as you. Recognizing closure rate early allows you to maintain your following distance without late braking.

HW- Judging Space

Essential Questions

What are the benefits of managing space and judging time, where is the point of no return and how does it impact our driving decisions, how do you create, maintain and/or rebuild sufficient following distance.

The amount of time/space between vehicles when following another vehicle in the intended path of travel to avoid conflict.

Limit your ability to gather information about the condition of your intended path.

Chance of injury, damage, or loss. In driving, risk (potential or immediate) is the possibility of having a conflict that results in a crash or collision.

Point beyond which a driver can no longer stop safely without entering the intersection or space.

ORPC - R3 Instructor Manual Copyright © 2022 by Western Oregon University. All Rights Reserved.

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