Kinematics Quest Answer key PDF

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Solutions for Kinematics Quest...

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jonnalagadda (saj2436) – Kinematics – ayida – (4355-3) This print-out should have 41 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Welcome to your first Quest assignment! This assignment will be due on the date and time posted on Quest. Remember, you do get some partial credit for at least attempting every single question, so even if you’re stuck, give it a shot. The grade you see when you finish the assignment will count as 1/2 of a test grade! 001 10.0 points The largest metric prefix that is ordinarily used is tera, while the smallest prefix is pico. How much larger is a terameter than a picometer? 1. They are different names for the same things. 2. 103 3. 1024 correct 4. None of these 5. 10

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6. 109 7. 1012 Explanation: A terameter is 1012 meters, while a picometer is 10−12 meters, so that the terameter is 1012 = 1024 as large as a picometer: −12 10   1012 m 1 picometer (1 terameter ) · terrameter 10−12 m 24 = 10 picometer . Exponents SUBTRACT in division: 1012 = 1012−(−12) = 1012+12 = 1024 . 10−12 002 (part 1 of 5) 10.0 points Use the SI prefixes to convert these hypothet-

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ical units of measure into appropriate quantities. Express 27 rations in dekarations. Correct answer: 2.7 dekaration. Explanation: The prefix “deka” is a variation of “deca”. 1 dekaration = 10 ration , so   1 dekaration 2.7 × 101 rations · 10 rations = 2.7 dekaration . 003 (part 2 of 5) 10.0 points Express 2164 mockingbirds in kilomockingbirds. Correct answer: 2.164 kmockingbird. Explanation: = 103 mockingbird , so  1 kmockingbird  2.164 × 103 mockingbirds 1 kmockingbird × 3 = 2.164 kmockingbird . 10 mockingbird 004 (part 3 of 5) 10.0 points WITHDRAWN 005 (part 4 of 5) 10.0 points WITHDRAWN 006 (part 5 of 5) 10.0 points WITHDRAWN 007 10.0 points Rain drops fall on a tile surface at a density of 4955 drops/ft2 . There are 17 tiles/ft2. How many drops fall on each tile? Correct answer: 291.471 drops/tile. Explanation: Let :

ρ = 4955 drops/ft2 n = 17 tiles/ft2 .

and

Applying dimensional analysis, R=

4955 drops/ft2 ρ = 17 tiles/ft2 n

= 291.471 drops/tile .

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3)

Correct answer: −78 km. Explanation:

8 6 4 2 0 −2 −4 −6 −8

−10

Let :

Displacement vs Time

10

displacement (m)

008 (part 1 of 2) 10.0 points While John is traveling along a straight interstate highway, he notices that the mile marker reads 253 km. John travels until he reaches the 132 km marker and then retraces his path to the 175 km marker. What is John’s resultant displacement from the 253 km marker?

s0 = 253 km and sf = 175 km .

0 2 4 6

8 10 12 14 16 18 20 time (s) What is the displacement at 10 s? 1. −1 m 2. None of these

∆s = sf − s0 = 175 km − 253 km = −78 km .

009 (part 2 of 2) 10.0 points How far has he traveled? Correct answer: 164 km. Explanation: Let :

2

3. Unable to determine 4. 1 m 5. −2 m 6. −3 m 7. 3 m 8. 2 m

s2 = 132 km .

The distance traveled is given by d = |s2 − s0| + |sf − s2| = |132 km − 253 km| + |175 km − 132 km| = 164 km .

9. −4 m correct 10. 0 m Explanation: Read the displacement from the graph. 011 (part 2 of 2) 10.0 points What is the velocity at 10 s? 1. −2 m/s 2. None of these

010 (part 1 of 2) 10.0 points Consider the following graph

3. 4 m/s

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3) 4. 2 m/s

4. 12.5 mph

5. Unable to determine

5. 25 mph

6. 3 m/s 7. 0 m/s correct

3

Explanation: The average velocity is final position − initial position final time − initial time 0 mi = 0 mph . = 8h

vav = 8. −1 m/s 9. −3 m/s 10. 1 m/s Explanation: At 10 s, the slope is 0. 012 10.0 points A particle accelerates from rest at 5.4 m/s2 . What is its speed 7.8 s after the particle starts moving? Correct answer: 42.12 m/s. Explanation: The initial speed vo = 0, so the speed of the particle after t seconds is v = v0 + a t = a t = (5.4 m/s2 )(7.8 s) = 42.12 m/s . 013 (part 1 of 3) 10.0 points Rabid Texas basketball fans have been known to rush out of their laboratory class at 5:00 p.m. and immediately get on highway I-35 to arrive in Waco, 100 miles away, at 7:00 p.m. to see the Baylor-Longhorn basketball game. They spend 4 hours watching and then celebrating the outcome of the game, and then drive 2 hours back to Austin. After they completed their trip, what is their mean velocity for their trip? mph is miles per hour 1. 0 mph correct 2. 100 mph 3. 50 mph

014 (part 2 of 3) 10.0 points After they completed their trip, what is their mean speed for the trip? 1. 25 mph correct 2. 50 mph 3. 100 mph 4. 12.5 mph 5. 0 mph Explanation: The average speed is 200 mi total distance = = 25 mph . time 8h 015 (part 3 of 3) 10.0 points What is their mean speed from when they leave Austin and first arrive in Waco? 1. 25 mph North 2. 50 mph South 3. 50 mph North 4. 25 mph 5. 25 mph South 6. 50 mph correct Explanation:

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3)

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Their mean speed is 100 mi total distance = 50 mph . = 2h time Speed is a scalar and does not have a direction.

v t

5.

016 (part 1 of 4) 10.0 points The following 4 questions refer to a toy car which can move to the right or left along a horizontal line. The positive direction is to the right. Choose the correct velocity-time graph for each of the following questions. Assume friction is so small that it can be ignored.

v t

6.

7. None of these graphs is correct. v

car

g

v

t

8. +

O

Which velocity graph shows the car moving toward the right (away from the origin) at a steady (constant) velocity?

correct v t

9.

v t

1.

v t

10.

v t

2.

Explanation: Since the velocity is constant, the graph is a straight line. Since the car is moving to the right the velocity is positive.

v t

3.

v t

v 4.

t

017 (part 2 of 4) 10.0 points Which velocity graph shows the car moving towards the right (away from the origin) reversing direction and then moving to the left?

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3)

5

v

v

t

9.

t

1.

v

v

t

10.

t

2.

v t

3.

Explanation: Since the car reverses its direction, the velocity is positive and then negative. v

v

t t

4.

018 (part 3 of 4) 10.0 points Which velocity graph shows the car moving toward the left (toward the origin) at a steady (constant) velocity?

5. None of these graphs is correct. v t

6.

v t

1. v t

7.

v t

2. correct v 8.

v t

3.

correct

t

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3) v

v

t

4.

t

2.

v

v

t

5.

t

3.

v

v

t

6.

t

4.

v

v

t

7.

t

5.

v

v

t

8.

6

t

6.

9. None of these graphs is correct. v v 10.

t

7. t

v Explanation: Constant velocity is a straight line; toward the left is negative velocity. 019 (part 4 of 4) 10.0 points Which velocity graph shows the car increasing its speed towards the right (away from the origin) at a steady (constant) rate? 1. None of these graphs is correct.

8.

t

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3) v

7

1. No t

9.

2. Yes correct 3. There is no way to tell from the information given.

correct v t

10.

Explanation: Since the car’s speed is increasing at a constant rate, the slope of the graph is a constant. v t

4. Acceleration is unrelated to speed, so the question does not make sense. Explanation: The acceleration is the rate of change of velocity. Since the object is moving and its velocity is changing, it is accelerating. 022

10.0 points

Consider a toy car which moves to the right (positive direction) on a horizontal surface along a straight line. car

020 10.0 points If the acceleration of an object is zero at some instant in time, what can be said about its velocity at that time?

v +

O

Which acceleration-time graph corresponds to the motion of the car if it moves toward the right, while slowing down at a steady rate.

1. It is negative. 2. Unable to determine.

a

3. It is zero.

t

1. 4. It is positive. 5. It is not changing at that time. correct a

Explanation: The acceleration a=

∆v =0 ∆t ∆v = 0 .

t

2.

a 021 10.0 points At a certain instant a moving object comes to momentary rest. Is it accelerating at that moment?

3.

t

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3) correct

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a a

t

c) t

4.

v t

d)

a t

5.

a 6. None of these graphs is correct.

e)

t

a t

7.

1. a), b), and c) 2. d) only 3. a) only

a t

8.

4. e) only 5. a) and d) correct 6. a) and c)

Explanation: Since the car slows down, the acceleration is in the opposite direction to the velocity.

7. a), b), and e)

023 10.0 points Identify all graphs that represent motion at constant speed (note the axes carefully).

9. a), b), and d)

x t

a)

v b)

t

8. None of these

10. c) only Explanation: For constant speed, a = 0 . a) x = k t, k > 0, increases at a constant rate, so it has a constant velocity. b) v = k t, k > 0, increases at a constant rate, so it has a constant acceleration. c) a = k t, k > 0, increases at a constant rate. d) v = k, k > 0, represents a constant velocity. e) a = k, k > 0, represents a constant,

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3) nonzero acceleration. 024 10.0 points Can an object reverse its direction of travel while maintaining a constant acceleration? 1. No; if the acceleration is constant, the direction of the speed remains unchanged. 2. Yes; a ball tossed upward reverses its direction of travel at its highest point. correct

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026 (part 1 of 3) 10.0 points A certain automobile manufacturer claims that its super-deluxe sports car will accelerate from rest to a speed of 41 m/s in 7.9 s. Find the acceleration of the car. Assume that the acceleration of the car is constant. Correct answer: 5.18987 m/s2 . Explanation: Let :

3. No; the direction of the speed is always the same as the direction of the acceleration.

vi = 0 , vf = 41 m/s , t = 7.9 s .

and

4. All are wrong. 5. Yes; a ball thrown toward a wall bounces back from the wall.

vf = vi + a t = a t vf 41 m/s a= = = 5.18987 m/s2 . t 7.9 s

Explanation: Velocity and acceleration need not be in the same direction. When a ball is tossed upward it experiences a constant acceleration directed downward.

027 (part 2 of 3) 10.0 points Find the distance the car travels in the first 7.9 s. Correct answer: 161.95 m.

025 10.0 points Which of the following is an example of something that undergoes acceleration while moving at constant speed? 1. A car moving straight backwards on the road 2. A football flying in the air 3. A man standing in an elevator 4. None of these. An object that undergoes an acceleration has to change its speed 5. A car making a circle in a parking lot correct Explanation: When an object moves in a circular path at constant speed its direction changes, so its velocity changes, meaning it experiences acceleration.

Explanation: Let :

xi = 0 .

xf = x0 + a t = a t = x=

1 (vf + vi ) t 2

1 (41 m/s) (7.9 s) = 161.95 m . 2

028 (part 3 of 3) 10.0 points What is the speed of the car 9.36 s after it begins its motion, assuming it continues to accelerate at the same average rate? Correct answer: 48.5772 m/s. Explanation: Let :

t = 9.36 s .

v = v0 + a t = a t = (5.18987 m/s2 ) (9.36 s) = 48.5772 m/s .

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3) 029 (part 1 of 2) 10.0 points A plane landing on a small tropical island has just 64 m of runway on which to stop. If its initial speed is 65 m/s, what is the maximum acceleration of the plane during landing, assuming it to be constant? Correct answer: −33.0078 m/s2 . Explanation:

Let :

s = 64 m and v0 = 65 m/s .

4. less than 9.8 m/s2 . Explanation: The acceleration due to the gravity is independent of any initial velocity and remains constant. 032 10.0 points An object is shot vertically upward into the air with a positive initial velocity. What correctly describes the velocity and acceleration of the object at its maximum elevation? Velocity

Acceleration

v 2 = v02 + 2 a s = 0 −v02 −(65 m/s)2 a= = 2(64 m) 2s

1. zero

negative correct

2. negative

negative

= −33.0078 m/s2 .

3. zero

zero

4. positive

negative

5. positive

positive

030 (part 2 of 2) 10.0 points How long does it take for the plane to stop with this acceleration? Correct answer: 1.96923 s. Explanation: v = v0 + a t = 0 −v0 −65 m/s t= = −33.0078 m/s2 a = 1.96923 s .

031 10.0 points If you drop an object, it will accelerate downward at a rate of g = 9.8 m/s2. If you throw it downward instead, its acceleration (in the absence of air resistance) will be 1. greater than 9.8 m/s2 . 2. Unable to determine. 2

3. 9.8 m/s correct

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Explanation: At the maximum elevation, the vertical velocity is zero. The acceleration is due to gravity, which always acts down. 033 (part 1 of 2) 10.0 points The tallest volcano in the solar system is the 18 km tall Martian volcano, Olympus Mons. An astronaut drops a ball off the rim of the crater and that the free fall acceleration of the ball remains constant throughout the ball’s 18 km fall at a value of 3.8 m/s2 . (We assume that the crater is as deep as the volcano is tall, which is not usually the case in nature.) Find the time for the ball to reach the crater floor. Correct answer: 97.3329 s. Explanation:

Let :

h = 18 km and a = 3.8 m/s2 .

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3) The distance an object falls from rest under an acceleration a is 1 2 at 2 s r 2 (18 km) 2h t= = = 97.3329 s . a (3.8 m/s2)

h=

034 (part 2 of 2) 10.0 points Find the magnitude of the velocity with which the ball hits the crater floor. Correct answer: 369.865 m/s. Explanation: Since the object falls from rest, v2 = 2 a h √ = q2 a h v=

2 (3.8 m/s2 ) (18 km)

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v02 correct 2g v2 4. hmax = √ 0 2g v02 5. hmax = g √ 2 3 v0 6. hmax = 2g 3 v2 7. hmax = 0 4g v2 8. hmax = 0 4g 5 v2 9. hmax = 0 8g Explanation: With up positive, a = −g and for the upward motion tf = 0, so for constant acceleration, 3. hmax =

= 369.865 m/s . 0 = v02 − 2 g (hmax − 0) v2 hmax = 0 . 2g 036 (part 1 of 2) 10.0 points A ball is thrown upward. After reaching a maximum height, it continues falling back toward Earth. On the way down, the ball is caught at the same height at which it was thrown upward.

v0

hmax

v0

hmax

035 10.0 points A ball is thrown upward with an initial vertical speed of v0 to a maximum height of hmax .

  vf2 = v02 + 2 a yf − y0

What is its maximum height hmax ? The acceleration of gravity is g. Neglect air resistance. √ 2 3v 1. hmax = √ 0 2 2g √ 2 5v 2. hmax = √ 0 2 2g

If the time (up and down) the ball remains in the air is 1.78 s, find its speed

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3) when it caught. The acceleration of gravity is 9.8 m/s2 . Neglect air resistance. Correct answer: 8.722 m/s. Explanation: Let :

vf = −v0 , ∆t = 1.78 s , and g = −9.8 m/s2 .

A daring stunt woman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 5.9 m/s, and the woman is initially 1.48 m above the level of the saddle. How long is she in the air? The acceleration of gravity is 9.8 m/s2 . Correct answer: 0.549582 s. Explanation:

Let : ∆y = −1.48 m and g = 9.8 m/s2 .

vf = v0 + g ∆t −v0 = v0 + g ∆t 2 v0 ∆t = − g g ∆t (−9.8 m/s2) (1.78 s) v0 = − =− 2 2 = 8.722 m/s .

For free fall, ∆y = t=

037 (part 2 of 2) 10.0 points If the time the ball remains in the air is 1.91 s, find the maximum height hmax the ball attained while in the air. Correct answer: 4.46892 m. Explanation: Consider the motion upward.

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1 2 gt 2 s

2 ∆y = g

s

2 (−1.48 m) −9.8 m/s2

= 0.549582 s .

039 (part 2 of 2) 10.0 points What must be the horizontal distance between the saddle and limb when the woman makes her move? Correct answer: 3.24254 m.

Let :

tup = 0.955 s and vf = 0 m/s .

Explanation: Let :

vf = v0 + g tup = 0 v0 = −g tup so 2 g tup 1 2 2 ∆y = v0 tup + g tup = −g t up + 2 2 2 g tup (−9.8 m/s2 ) (1.91 s)2 =− =− 2 2 = 4.46892 m .

038 (part 1 of 2) 10.0 points

vx = 5.9 m/s .

The horse moves with a constant velocity, so x = vx t = (5.9 m/s) (0.549582 s) = 3.24254 m .

040 10.0 points A stone is thrown straight upward and at the top of its trajectory its velocity is momentarily zero.

jonnalagadda (saj2436) – Kinematics – ayida – (4355-3)

What is its acceleration at this point? 1. 9.8 m/s2 down correct 2. Unable to determine 3. 9.8 m/s2 up 4. Zero Explanation: The gravitational acceleration near the surface of the earth is considered constant for all practical purposes. This acceleration of 9.8 m/s2 points downward. 041 10.0 points The most common frame of reference is 1. the object you are observing. 2. the sky. 3. the Earth. correct 4. you. Explanation:

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