rogers (nr9358) – Kinematics #1 – volle – (301) 1

This print-out should have 21 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 (part 1 of 2) 10.0 pointsWhile John is traveling along a straight inter-state highway, he notices that the mile markerreads 256 km. John travels until he reachesthe 144 km marker and then retraces his pathto the 171 km marker.What is John’s resultant displacement from

the 256 km marker?

Correct answer: −85 km.

Explanation:

Let : s0 = 256 km and

sf = 171 km .

∆s = sf − s0 = 171 km− 256 km

= −85 km .

002 (part 2 of 2) 10.0 pointsHow far has he traveled?

Correct answer: 139 km.

Explanation:

Let : s2 = 144 km .

The distance traveled is given by

d = |s2 − s0|+ |sf − s2|= |144 km− 256 km|

+ |171 km− 144 km|= 139 km .

003 (part 1 of 4) 10.0 pointsBoth car A and car B leave school at the sametime, traveling in the same direction. Car Atravels at a constant speed of 79 km/h, whilecar B travels at a constant speed of 87 km/h.

How far is Car A from school 1.8 h later?

Correct answer: 142.2 km.

Explanation:

Let : va = 79 km/h and

t = 1.8 h .

d = va t = (79 km/h) (1.8 h) = 142.2 km .

004 (part 2 of 4) 10.0 pointsHow far is car B from school 1.8 h later?

Correct answer: 156.6 km.

Explanation:

Let : vb = 87 km/h .

d = vb t = (87 km/h) (1.8 h) = 156.6 km .

005 (part 3 of 4) 10.0 pointsHow long does it take car A to reach a gasstation a distance 174 km from the school?

Correct answer: 2.20253 h.

Explanation:

Let : d = 174 km .

t =d

va=

174 km

79 km/h= 2.20253 h .

006 (part 4 of 4) 10.0 pointsHow long does it take car B to reach the gasstation?

Correct answer: 2 h.

Explanation:

t =d

vb=

174 km

87 km/h= 2 h .

rogers (nr9358) – Kinematics #1 – volle – (301) 2

007 10.0 pointsA glacier moves with a speed of 54 nm/s.How many years would it take for the

glacier to move 0.65 km? Assume there are365 days in a year.

Correct answer: 381.692 y.

Explanation:

Let : d = 0.65 km = 650 m and

v = 54 nm/s = 5.4× 10−8 m/s .

The time is

t =d

v=

650 m

5.4× 10−8 m/s

1 h

3600 s

× 1 d

24 h

1 y

365 d

= 381.692 y .

008 10.0 pointsAn oceanic depth-sounding vessel surveys theocean bottom with ultrasonic waves thattravel 1530 m/s in seawater.How deep is the water directly below the

vessel if the time delay of the echo to theocean floor and back is 6 s?

Correct answer: 4590 m.

Explanation:

Let : v = 1530 m/s and

∆ t = 6 s .

The sound takes 3 s to reach the ocean floor(and 3 s to return), so

d = v t = (1530 m/s) (3 s) = 4590 m .

009 (part 1 of 2) 10.0 pointsA cyclist maintains a constant velocity of6.7 m/s headed away from point A. At someinitial time, the cyclist is 249 m from point A.What will be his displacement from his

starting position after 62 s?

Correct answer: 415.4 m.

Explanation:

Let : v = 6.7 m/s and

t = 62 s .

The distance is

∆d = v t = (6.7 m/s) (62 s) = 415.4 m .

010 (part 2 of 2) 10.0 pointsWhat will be his position from point A afterthat time?

Correct answer: 664.4 m.

Explanation:

Let : d0 = 249 m .

The position is

d = d0 +∆d = 249 m+ 415.4 m = 664.4 m .

011 (part 1 of 2) 10.0 pointsThe position versus time for a certain objectmoving along the x-axis is shown. The ob-ject’s initial position is 1 m.

−6

−4

−2

0

2

4

6

8

0 1 2 3 4 5 6 7 8 9

b

b

b b

b

b

time (s)

position(m

)

Find the instantaneous velocity at 6.5 s.

Correct answer: −2.33333 m/s.

Explanation:

rogers (nr9358) – Kinematics #1 – volle – (301) 3

v =−4 m− (3 m)

8 s− 5 s= −2.33333 m/s .

012 (part 2 of 2) 10.0 pointsFind the average velocity over the whole timeshown.

Correct answer: 0.333333 m/s.

Explanation:

v =4 m− (1 m)

9 s− 0 s= 0.333333 m/s .

013 (part 1 of 4) 10.0 pointsThe position versus time for a certain objectmoving along the x-axis is shown. The ob-ject’s initial position is −4 m.

−6

−4

−2

0

2

4

6

8

10

0 1 2 3 4 5 6 7 8 9

b

b

b b

b

b

time (s)

position(m

)

Find the instantaneous velocity at 1 s.

Correct answer: 6 m/s.

Explanation:The instantaneous velocity is the slope of

the tangent line at that point.

v =8 m− (−4 m)

2 s− 0 s= 6 m/s .

014 (part 2 of 4) 10.0 pointsFind the instantaneous velocity at 2.5 s.

Correct answer: −4 m/s.

Explanation:

v =4 m− (8 m)

3 s− 2 s= −4 m/s .

015 (part 3 of 4) 10.0 pointsFind the instantaneous velocity at 4 s.

Correct answer: 0 m/s.

Explanation:

v =4 m− (4 m)

5 s− 3 s= 0 m/s .

016 (part 4 of 4) 10.0 pointsFind the instantaneous velocity at 8.5 s.

Correct answer: 4 m/s.

Explanation:

v =0 m− (−4 m)

9 s− 8 s= 4 m/s .

017 (part 1 of 5) 10.0 pointsConsider the displacement curve OABC

0 1 2−2

−1

0

1

2

x(m

)

t (s)

Displacement vs Time

O

A

B

C

What is the average velocity from point Oto A?

1. vOA

= −√3 m/s

2. vOA

= +√3 m/s

3. vOA

= −2 m/s

rogers (nr9358) – Kinematics #1 – volle – (301) 4

4. vOA

= 0 m/s

5. vOA

= +2 m/s correct

Explanation:The displacement x is on the vertical axis

and the time t is on the horizontal axis. Ve-locity requires net displacement:

vOA

=x

A− x

O

tA− t

O

=2− 0

1− 0= +2 m/s .

018 (part 2 of 5) 10.0 pointsWhat is the average velocity for the motionfrom point O to point B?

1. vOB

= +√3 m/s

2. vOB

= −√3 m/s

3. vOB

= 0 m/s correct

4. vOB

= +2 m/s

5. vOB

= −2 m/s

Explanation:

vOB

=x

B− x

O

tB− t

O

=0− 0

2− 0= 0 m/s .

019 (part 3 of 5) 10.0 pointsWhat is the average speed for the motion frompoint O to point B?

1. sOB

= −2 m/s

2. sOB

= +2 m/s correct

3. sOB

= −√3 m/s

4. sOB

= 0 m/s

5. sOB

= +√3 m/s

Explanation:

sOB

=|x

A− x

O|+ |x

B− x

A|

tB− t

O

=2 + 2

2− 0= +2 m/s .

020 (part 4 of 5) 10.0 pointsWhat is the instantaneous velocity at pointB?

1. vB= −2 m/s correct

2. vB= +2 m/s

3. vB= +

√3 m/s

4. vB= 0 m/s

5. vB= −

√3 m/s

Explanation:The instantaneous velocity at point B can

be obtained by first finding an expression forthe position of the moving object and takingits derivative evaluated at time tB. Sincethe graph near B is linear, it is simpler tocalculate the slope of the line over the intervalfrom A to B. The result correctly describesthe instantaneous velocity at B because thederivative of a straight line is constant at allpoints on the line and can be obtained for ourcase by

vB=

xB− x

A

tB− t

A

=0− 2

2− 1= −2 m/s .

021 (part 5 of 5) 10.0 pointsWhat is the instantaneous speed at point B?

1. sB= −

√3 m/s

2. sB= +2 m/s correct

3. sB= −2 m/s

4. sB= 0

5. sB= +

√3 m/s

Explanation:Instantaneous speed is simply the magni-

tude of the velocity:

sB= |v

B| = | − 2| = +2 m/s .