Version 001 Summer Review #03 tubman (IBII ) 1

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This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.

Concept 20 P03 001 10.0 points

An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic waves that travel 1530 m/s in seawater.

How deep is the water directly below the vessel if the time delay of the echo to the ocean floor and back is 4 s?

Correct answer: 3060 m. Explanation:

Let : v = 1530 m/s and ∆ t = 4 s .

The sound takes 2 s to reach the ocean floor (and 2 s to return), so

d = v t = (1530 m/s) (2 s) = 3060 m .

Constant Velocity 02 002 10.0 points

An object is moving with constant velocity ~v. Which situation is impossible in such a cir-cumstance?

1. Four forces act on the object.

2. One force acts on the object. correct 3. Three forces act on the object.

4. Two forces act on the object. 5. No forces act on the object. Explanation:

Two or more vectors can sum to zero, but a single nonzero vector cannot sum to zero, and Newton’s Second Law requires that

X ~ F = m~a = 0 in this case. Displacement Curve 02 003 (part 1 of 2) 10.0 points

Consider a moving object whose position x is plotted as a function of the time t. The object moved in different ways during the time intervals denoted I, II and III on the figure. 2 4 6 2 4 6 t x I II III

During these three intervals, when was the object’s speed highest? Do not confuse the speed with the velocity.

1. Same speed during each of the three in-tervals.

2. During interval I 3. During interval II

4. Same speed during intervals II and III 5. During interval III correct

Explanation:

The velocity v is the slope of the x(t) curve; the magnitude v = |v| of this slope is the speed. The curve is steepest (in absolute magnitude) during the interval III and that is when the object had the highest speed.

004 (part 2 of 2) 10.0 points

During which interval(s) did the object’s ve-locity remain constant?

1. During none of the three intervals 2. During interval II only

3. During each of the three intervals cor-rect

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4. During interval III only 5. During interval I only Explanation:

For each of the three intervals I, II or III, the x(t) curve is linear, so its slope (the velocity v) is constant. Between the intervals, the velocity changed in an abrupt manner, but it did remain constant during each interval.

Average Speed on a Trip 005 (part 1 of 2) 10.0 points

A person travels by car from one city to an-other. She drives for 20.9 min at 78 km/h, 8.05 min at 114 km/h, 39.9 min at 48.3 km/h, and spends 17.8 min along the way eating lunch and buying gas.

Determine the distance between the cities along this route.

Correct answer: 74.5845 km. Explanation: Let : t1= 20.9 min , v1= 78 km/h , t2= 8.05 min , v2= 114 km/h , t3= 39.9 min , and v3= 48.3 km/h . x = x1+ x2+ x3 = v1t1+ v2t2+ v3t3 = (78 km/h)(20.9 min) + (114 km/h)(8.05 min) + (48.3 km/h)(39.9 min) = 74.5845 km . 006 (part 2 of 2) 10.0 points Determine the average speed for the trip. Correct answer: 51.6454 km/h.

Explanation:

Let : tother = 17.8 min .

The total time is

t = t1+ t2+ t3+ t_other = 20.9 min + 8.05 min + 39.9 min + 17.8 min = 1.44417 h , so vav = x t = 74.5845 km 1.44417 h = 51.6454 km/h . Comparison of Average Velocity

007 10.0 points

The position-versus-time graph below de-scribes the motion of three different bodies (labeled 1, 2, 3). 1 2 3 x_B x_A t_A t_B x t A B

Consider the average velocities of the three bodies. Which of the following statements is correct? 1. ¯v1 > ¯v2 > ¯v3 2. ¯v1 = ¯v2 = ¯v3 correct 3. ¯v1 < ¯v2 < ¯v3 4. ¯v1 > ¯v2 and ¯v3 > ¯v2 Explanation: Average velocity is ¯ v = displacement time = x_B _{− xA} t_B _{− tA} .

All three bodies have exactly same displace-ment in exactly same time, so the average velocities are exactly equal: ¯v1 = ¯v2= ¯v3.

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Accelerating Object 008 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?

1. It is positive.

2. It is not changing at that time. correct 3. Unable to determine. 4. It is negative. 5. It is zero. Explanation: The acceleration a = ∆v ∆t = 0 ∆v = 0 .

Acceleration Time Graph 01 009 (part 1 of 5) 10.0 points

Consider a toy car which can move to the right (positive direction) or left on a horizon-tal surface along a straight line.

car

v

O +

What is the acceleration-time graph if the car moves toward the right (away from the origin), speeding up at a steady rate?

1. t

a

2. None of these graphs is correct.

3. t a correct 4. t a 5. t a 6. t a 7. t a 8. t a Explanation:

Since the car speeds up at a steady rate, the acceleration is a constant.

What is the acceleration-time graph if the car moves toward the right, slowing down at a steady rate?

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1. t a 2. t a 3. t a 4. t a correct 5. t a 6. t a 7. t a

8. None of these graphs is correct. Explanation:

Since the car slows down, the acceleration is in the opposite direction.

What is the acceleration-time graph if the car moves towards the left (toward the origin) at a constant velocity? 1. t a 2. t a 3. t a correct

5. t a 6. t a 7. t a

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8. t a

Explanation:

Since the car moves at a constant velocity, the acceleration is zero.

What is the acceleration-time graph if the car moves toward the left, speeding up at a steady rate? 1. t a 2. t a correct 3. t a 4. t a

6. t a 7. t a 8. t a Explanation:

The same reason as Part 1.

What is the acceleration-time graph if the car moves toward the right at a constant velocity?

1. t a 2. t a 3. t a 4. t a

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5. t a

correct

6. t

a

8. t

a

Explanation:

The same reason as Part 3.

Acceleration of a Vehicle 014 10.0 points

A car traveling in a straight line has a velocity of 2.34 m/s at some instant. After 5.46 s, its velocity is 12.2 m/s.

What is its average acceleration in this time interval? Correct answer: 1.80586 m/s2. Explanation: Let : v1 = 2.34 m/s , t = 5.46 s , and v2 = 12.2 m/s .

The average acceleration is aav = ∆v ∆t = vf − vi ∆t = 12.2 m/s − 2.34 m/s 5.46 s = 1.80586 m/s2 . Acceleration vs Time 01 015 (part 1 of 4) 10.0 points

Consider the plot below describing the accel-eration of a particle along a straight line with an initial position of 25 m and an initial ve-locity of −8 m/s. −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 0 1 2 3 4 5 6 7 8 9 time (s) a cc el er a ti o n (m / s 2 )

What is the velocity at 1 s? Correct answer: −5 m/s. Explanation: Let : v0 = −8 m/s , a1 = 3 m/s 2 , and ∆t = 1 s .

The change in velocity is the area a ∆t be-tween the acceleration curve and the time axis.

vf = v0+ a1∆t

= −8 m/s + (3 m/s2) (1 s) = _{−5 m/s .}

016 (part 2 of 4) 10.0 points What is the position at 1 s?

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Correct answer: 18.5 m. Explanation: Let : x0 = 25 m . x1 = x0+ v0∆t + 1 2a1∆t 2 = 25 m + (−8 m/s) (1 s) +1 2(3 m/s 2 ) (1 s)2= 18.5 m . 017 (part 3 of 4) 10.0 points What is the velocity at 7 s?

Correct answer: 3 m/s. Explanation: Let : ∆t1 = 6 s , a2 = −7 m/s 2 , and ∆t2 = 1 s .

The calculation is done in two parts, each with constant acceleration.

The velocity after 6 s is v1= v0+ a1∆t1

= −8 m/s + (3 m/s2)(6 s) = 10 m/s , so the velocity after 7 s is

v2 = v1+ a2∆t2

= 10 m/s + (−7 m/s2)(1 s) = 3 m/s .

018 (part 4 of 4) 10.0 points What is the position at 7 s?

Correct answer: 37.5 m. Explanation:

The position after 6 s is

x1 = x0+ v0∆t1+ 1 2 a1(∆t1) 2 = (25 m) + (−8 m/s) (6 s) +1 2(3 m/s 2 ) (6 s)2= 31 m ,

so the position after 7 s is

x2 = x1+ v1∆t2+ 1 2 a2(∆t2) 2 = 31 m + (10 m/s) (1 s) +1 2 (−7 m/s 2 ) (1 s)2= 37.5 m . Describing Motion 019 (part 1 of 2) 10.0 points

A car initially at rest on a straight road accelerates according to the acceleration vs time plot.

t a

t1 t2

t3 t4

What is the best description of the motion of the car? Take forward to be the positive direction.

1. The car starts at rest, accelerates to a low speed, comes to a stop, accelerates backwards and cruises in reverse.

2. The car goes forward and then goes back-ward, ending where it started.

3. The car starts at rest, accelerates to a high speed, cruises for a short while, decelerates to a lower speed, then cruises. correct

4. The car goes backward and then goes forward.

5. The car goes forward and then goes back-ward, ending behind where it started.

6. The car starts at rest, goes up to a high speed, stops moving, travels backward, and stops.

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locity behavior: 1) v0 = 0 (initially at rest) 2) 0 < t < t1 : a = 0 (remains at rest) 3) t1 < t < t2 : a > 0 (accelerates forward) 4) t2 < t < t3 : a = 0 (constant speed) 5) t3 < t < t4 : a < 0 (decelerates) 6) t4 < t : a = 0 (constant speed) 020 (part 2 of 2) 10.0 points

Which of the following graphs describes the velocity vs time of the car?

1. t

v

t1 t2

t3 t4

3. t v t1 t2 t3 t4 4. t v t1 t2 t3 t4 5. t v t1 t2 t3 t4 6. t v t1 t2 t3 t4 7. t v t1 t2 t3 t4 8. t v t1 t2 t3 t4 9. t v t1 t2 t3 t4 correct Explanation: 1) 0 < t < t1 : v0 = 0

2) t1 < t < t2 : velocity increases uniformly

3) t2 < t < t3 : constant velocity

4) t3 < t < t4 : velocity decreases uniformly

5) t4 < t : constant velocity

The velocity vs time graph can be found by plotting the area under the acceleration vs time curve.

Hewitt CP9 03 E07 021 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 di-rection of travel at its highest point. correct

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

4. Yes; a ball thrown toward a wall bounces back from the wall.

5. All are wrong. Explanation:

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same direction. When a ball is tossed upward it experiences a constant acceleration directed downward.

Hewitt CP9 03 E13 022 10.0 points

Which of the following is an example of some-thing that undergoes acceleration while mov-ing at constant speed?

1. None of these. An object that undergoes an acceleration has to change its speed

2. A car moving straight backwards on the road

3. A football flying in the air 4. A man standing in an elevator

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.

Velocity vs Time 08 023 (part 1 of 5) 10.0 points

The scale on the horizontal axis is 5 s and on the vertical axis 5 m/s. The initial position is 14 m. 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 8 9 time (× 5 s) v el o ci ty (× 5 m / s)

What is the initial velocity? Correct answer: 0.

Explanation:

At time t = 0, the velocity is 0. 024 (part 2 of 5) 10.0 points What is the final velocity?

Correct answer: 25 m/s. Explanation:

The final velocity is read from the graph and is 5 times the vertical scale factor (5 m/s).

vf = 5 (5 m/s) = 25 m/s .

025 (part 3 of 5) 10.0 points What is the final position?

Correct answer: 576.5 m. Explanation:

The change in position is the area under the velocity vs time graph, so

pf = po+ 1 2 (9 t) (5 v) = po+ 1 2 9 (5 s) 5 (5 m/s) = 576.5 m . 026 (part 4 of 5) 10.0 points

What acceleration is represented by the graph?

Correct answer: 0.555556 m/s2. Explanation:

The acceleration is the slope of the veloc-ity vs time graph, so

a = ∆v ∆t = 5 v − 0 9 t − 0 = 5 (5 m/s) 9 (5 s) = 0.555556 m/s2 . 027 (part 5 of 5) 10.0 points In which direction is the motion?

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2. forward correct 3. backward

Explanation:

The velocity is positive, so the motion is directed forward.

AP B 1993 MC 5 028 10.0 points

An object is released from rest on a planet that has no atmosphere. The object falls freely for 3 m in the first second.

What is the magnitude of the acceleration due to gravity on the planet?

1. 3.0 m/s2 2. 12.0 m/s2 3. 6.0 m/s2 correct 4. 10.0 m/s2 5. 1.5 m/s2 Explanation: Let : s = 3 m . s = 1 2a t 2 a = 2 s t2 = 2 (3 m) (1.0 s)2 = 6 m/s 2 . AP M 1993 MC 19 029 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

1. zero zero

2. zero negative correct

3. negative negative 4. positive positive 5. positive negative Explanation:

At the maximum elevation, the vertical ve-locity is zero. The acceleration is due to gravity, which always acts down.

Acceleration of Falling Object 030 10.0 points

If you drop an object, it accelerates downward at 9.8 m/s2 (in the absence of air resistance).

If, instead, you throw it downward, its downward acceleration after release is

1. 9.8 m/s2. correct 2. less than 9.8 m/s2. 3. more than 9.8 m/s2. Explanation:

The acceleration of all freely falling objects is g = 9.8 m/s2.

Ball M 01 031 10.0 points

A ball is thrown upward with an initial verti-cal speed of v0 to a maximum height of h_max.

b b b b b b b bb bb b b b b b b b b b v0 hm a x

What is its maximum height hmax? The

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resis-tance. 1. hmax = v02 4 g 2. hmax = 3 v2 0 4 g 3. hmax = √ 5 v2 0 2√2 g 4. hmax = v2 0 √ 2 g 5. hmax = v 2 0 g 6. hmax = √ 3 v02 2 g 7. hmax = 5 v2 0 8 g 8. hmax = √ 3 v2 0 2√2 g 9. hmax = v2 0 2 g correct Explanation:

With up positive, a = −g and for the up-ward motion tf = 0, so for constant

accelera-tion, v2_f = v02+ 2 a yf − y0 0 = v02− 2 g (hmax− 0) hmax = v 2 0 2 g.

Ball Drop on Olympus 032 (part 1 of 2) 10.0 points

The tallest volcano in the solar system is the 19 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 19 km fall at a value of 3.1 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: 110.716 s.

Explanation:

Let : h = 19 km and a = 3.1 m/s2.

The distance an object falls from rest under an acceleration a is h = 1 2 a t 2 t =r 2 h a = s 2 (19 km) (3.1 m/s2₎ = 110.716 s . 033 (part 2 of 2) 10.0 points

Find the magnitude of the velocity with which the ball hits the crater floor.

Correct answer: 343.22 m/s. Explanation:

Since the object falls from rest, v2= 2 a h =√2 a h v = q 2 (3.1 m/s2 ) (19 km) = 343.22 m/s . Holt SF 02F 03 034 (part 1 of 2) 10.0 points

A tennis ball is thrown vertically upward with an initial velocity of +7.0 m/s.

What will the ball’s velocity be when it returns to its starting point? The acceleration of gravity is 9.81 m/s2. Correct answer: −7 m/s. Explanation: Let : vi= 7.0 m/s , a = −9.81 m/s2, and ∆y = 0 m . v2_f = vi2+ 2 a ∆y = v 2 i vf = ±vi = ±7 m/s .

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is −7 m/s .

How long will the ball take to reach its starting point? Correct answer: 1.42712 s. Explanation: vf = vi+ a ∆t vf − vi = a ∆t ∆t = vf − vi a = −7 m/s − (7 m/s) −9.81 m/s2 = 1.42712 s .