SOLUTION MANUAL FUNDAMENTALS OF
PHYSICS, EXTENDED 12TH EDITION BY DAVID
HALLIDAY (AUTHOR), ROBERT RESNICK
(AUTHOR), JEARL WALKER (AUTHOR)-
QUESTIONS WITH DEATAILED SOLUTIONS| A+
,1. The Speed (Assumed Constant) Is (90 Km/H)(1000 M/Km) (3600 S/H) = 25 M/S.
Thus, During 0.50 S, The Car Travels (0.50)(25) 13 M.
,2. Huber’s Speed Is
V0=(200 M)/(6.509 S)=30.72 M/S = 110.6 Km/H,
Where We Have Used The Conversion Factor 1 M/S = 3.6 Km/H. Since Whittingham
Beat Huber By 19.0 Km/H, His Speed Is V1=(110.6 + 19.0)=129.6 Km/H, Or 36 M/S (1
Km/H = 0.2778 M/S). Thus, The Time Through A Distance Of 200 M For Whittingham
Is
X 200 M
T = = = 5.554 S.
V1 36 M/S
, 3. We Use Eq. 2-2 And Eq. 2-3. During A Time Tc When The Velocity Remains A
Positive Constant, Speed Is Equivalent To Velocity, And Distance Is Equivalent To
Displacement, With X = V Tc.
(a) During The First Part Of The Motion, The Displacement Is X1 = 40 Km And The
Time Interval Is
(40 Km)
T1 = = 1.33 H.
(30 Km /
H)
During The Second Part The Displacement Is X2 = 40 Km And The Time Interval Is
(40 Km)
T2 = = 0.67 H.
(60 Km /
H)
Both Displacements Are In The Same Direction, So The Total Displacement Is
X = X1 + X2 = 40 Km + 40 Km = 80 Km.
The Total Time For The Trip Is T = T1 + T2 = 2.00 H. Consequently, The Average Velocity
Is
(80 Km)
Vavg = = 40 Km / H.
(2.0 H)
(b) In This Example, The Numerical Result For The Average Speed Is The Same As
The Average Velocity 40 Km/H.
(c) As Shown Below, The Graph Consists Of Two Contiguous Line Segments, The First
Having A Slope Of 30 Km/H And Connecting The Origin To (T1, X1) = (1.33 H, 40
Km) And The Second Having A Slope Of 60 Km/H And Connecting (T1, X1) To (T, X) =
(2.00 H, 80 Km). From The Graphical Point Of View , The Slope Of The Dashed
Line Drawn From The Origin To (T, X) Represents The Average Velocity.
PHYSICS, EXTENDED 12TH EDITION BY DAVID
HALLIDAY (AUTHOR), ROBERT RESNICK
(AUTHOR), JEARL WALKER (AUTHOR)-
QUESTIONS WITH DEATAILED SOLUTIONS| A+
,1. The Speed (Assumed Constant) Is (90 Km/H)(1000 M/Km) (3600 S/H) = 25 M/S.
Thus, During 0.50 S, The Car Travels (0.50)(25) 13 M.
,2. Huber’s Speed Is
V0=(200 M)/(6.509 S)=30.72 M/S = 110.6 Km/H,
Where We Have Used The Conversion Factor 1 M/S = 3.6 Km/H. Since Whittingham
Beat Huber By 19.0 Km/H, His Speed Is V1=(110.6 + 19.0)=129.6 Km/H, Or 36 M/S (1
Km/H = 0.2778 M/S). Thus, The Time Through A Distance Of 200 M For Whittingham
Is
X 200 M
T = = = 5.554 S.
V1 36 M/S
, 3. We Use Eq. 2-2 And Eq. 2-3. During A Time Tc When The Velocity Remains A
Positive Constant, Speed Is Equivalent To Velocity, And Distance Is Equivalent To
Displacement, With X = V Tc.
(a) During The First Part Of The Motion, The Displacement Is X1 = 40 Km And The
Time Interval Is
(40 Km)
T1 = = 1.33 H.
(30 Km /
H)
During The Second Part The Displacement Is X2 = 40 Km And The Time Interval Is
(40 Km)
T2 = = 0.67 H.
(60 Km /
H)
Both Displacements Are In The Same Direction, So The Total Displacement Is
X = X1 + X2 = 40 Km + 40 Km = 80 Km.
The Total Time For The Trip Is T = T1 + T2 = 2.00 H. Consequently, The Average Velocity
Is
(80 Km)
Vavg = = 40 Km / H.
(2.0 H)
(b) In This Example, The Numerical Result For The Average Speed Is The Same As
The Average Velocity 40 Km/H.
(c) As Shown Below, The Graph Consists Of Two Contiguous Line Segments, The First
Having A Slope Of 30 Km/H And Connecting The Origin To (T1, X1) = (1.33 H, 40
Km) And The Second Having A Slope Of 60 Km/H And Connecting (T1, X1) To (T, X) =
(2.00 H, 80 Km). From The Graphical Point Of View , The Slope Of The Dashed
Line Drawn From The Origin To (T, X) Represents The Average Velocity.
