solutions manual for university physics with modern physics 2nd edition by wolfgang bauer and gary dwestfall
solutions manual for university physics with modern physics 2nd edition
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Solutions Manual for University Physics with Modern Physics
2nd Edition by Wolfgang Bauer and Gary D.Westfall
Chapter 2: Motion in a Straight Line
Chapter 2. Motion in a Straight Line
Concept Checks
2.1. d 2.2. b 2.3. b 2.4. c 2.5. a) 3 b) 1 c) 4 d) 22.6. c 2.7. d 2.8. c 2.9. d
Multiple-Choice Questions
2.1. e 2.2. c 2.3. c 2.4. b 2.5. e 2.6. a 2.7. d 2.8. c 2.9. a 2.10. b 2.11. b 2.12. d 2.13. c 2.14. d 2.15. a 2.16. c
Conceptual Questions
Velocity and speed are defined differently. The magnitude of average velocity and average speed are the same
only when the direction of movement does not change. If the direction cehsadnugring movement, it is known
that the net displacement is smaller than the net distance. Using the definition of average velocity and
speed, it can be said that the magnitude of average velocity is less than the average speed when the
direction changesduring movement. Here, only Christine changes direction during her movement.
Therefore, only Christine has a magnitude of average velocity which is smaller than her average speed.
The acceleration due to gravity is always pointing downward towards the center of the Earth.
It can be seen that the direction of velocity is opposite to the direction of acceleration when the ball is in
flight upward. The direction of velocity is the same as the direction of acceleration when the ball is in flight
downward.
The car, before the brakes are applied, has a constant velovc0i,tya,nd zero acceleration. After the brakes are
applied, the acceleration is constant and in the direction opposite to the velocity. In velocity versus time and
acceleration versus time graphs, the motion is described in the figures below.
There are two cars, car 1 and car 2. The decelerations ar1e= 2a2 = −a0 after applying the brakes. Before
applying the brakes, the velocities of both cars are the savm
1 =e,v2 v0=. When the cars have completely
45
,Solutions Manual for University Physics with Modern Physics
2nd Edition by Wolfgang Bauer and Gary D.Westfall
Bauer/WestfallU
: niversity Physics, 2E
v0 . Therefore, the ratio of time taken
stopped, the final velocities are zevrof =, 0 . vf = +v0 at= ⇒0 −=
ta
−v0 /−a0 1. So the ratio is one half. time
to stop is Ratio =time of car 1= =
of car 2 −v0/ 1 a0 2
2
Herea andv are instantaneous acceleration and velocitya. I=f 0 andv ≠ 0 at timet, then at that moment the object is
moving at a constant velocity. In other words, the slope of a curve in a velocity versus time plot iszero at
timet. See the plots below.
The direction of motion is determinedby the direction of velocity. Acceleration is defined as a change in velocity
per change in time. The change in veloci∆ tyv, , can be positive or negative depending on the values of initial
and final velocities∆
, =v vf −vi . If the acceleration is in the opposite direction to the motion, it means that
the magnitude of the objects velocity is decreasing. This occurs when an object is slowing down.
If there is no air resistance, then the acceleration does neoptendd on the mass of an object. Therefore, both
snowballs have the same acceleration. Since initial velocities are zero, and the snowballs will cover the same
distance, both snowballs will hit the ground at the same time. They will both have the sam.e speed
Acceleration is independent of the mass of an object if there is no air resistance.
Snowball 1 will return to its original position af∆
tetr , and then it falls in the same way as snowball 2.
Therefore snowball 2 will hit the ground first since it has a shorter path. However, both snowballs have the
same speed when they hit the ground.
46
,Solutions Manual for University Physics with Modern Physics
2nd Edition by Wolfgang Bauer and Gary D.Westfall
Chapter 2: Motion in a Straight Line
Make sure the scale for the dpilsacements of the car is correct. The length of the car is 174.9 in = 4.442 m.
Measuring the length of the car in the figure above with a ruler, the car in this scale is 0.80 ± 0.05 cm. Drawvertica
lines at the center of the car as shown in thuerfeigabove. Assume line 7 is the origixn=( 0).
Assume a constant acceleratioan= a0 . Use the equationsv = v0 at+ andx = x0 + v t0 + (1/2)at2 . When the
car has completely stoppedv,= 0at t =t0 .
0 =+v0 at⇒0 −= v0 at0
Use the final stopping position as the origxin=, 0at t =t0 .
0 = x0+ v t0 0+ 1 at02
2
Substitutingv0 = −at0 and simplifying gives
x0 −at02+ 12at02= 0 ⇒ x0 − 12at02= 0 ⇒ a= 2tx020
47
,Solutions Manual for University Physics with Modern Physics
2nd Edition by Wolfgang Bauer and Gary D.Westfall
Bauer/WestfallU
: niversity Physics, 2E
Note that timet0 is the time required to stop from a distanxc0e.First measure the length of the car. The length
of the car is 0.80 cm. The actual length of the car is 4.442 m, therefore the scale is
4.442 m= 0.275 m(roundat the
end). 0.80 cm
So the scale is 5.5 ± 0.275 m/cm. The farthest distance of the car from the origin is 2.9 ± 0.05 cm. Multiplying
by the scale, 15.95 mt0,= (0.333)(6 s) =1.998 s . The acceleration can be found usian=g 2x0
/t02 : a = 7.991 m/2s=. Because the scale has two significant digits, round the result to two
significant digitsa: = 8.0 m/s 2. Since the error in the measurement∆isx0 = 0.275 m, the error of the acceleration
is
∆ =a 2∆x0 2 0.275( = m) 0.1 m/s2.≈
t
Velocity can be estimated by computing the slope of a curve in a distance versus tim
. e plot
v v v
Velocity is defined byv∆ x∆/ t. If acceleration is constant, thean= f − i ∆ . (a) Estimate the slope
tf −ti ∆t of the dashed blue
line. Pick two points: it is more accurate to pick a point that coincides with horizontal lines of the grid.
Choosing pointst = 0 s,x= 0 m andt = 6.25 s,x = 20 m:v = 20. m−0 m = 3.2 m/s
6.25s −0 s
(b) Examine the sketch. There is a tangent to the curve t=
at7.5 s. Pick two points on the line. Choosing
points:t = 3.4 s,x= 0 m andt = 9.8 s,x= 60 m:
v= 60. m−0 m =
9.4 m/s9.8 s−3.4 s
(c) From (a),v = 3.2 m/s at = 2.5 s and from (b)v, = 9.4 m/s at = 7.5 s. From the definition of constant
acceleration,
a = 9.4 m/s−3.2 m/s= 6.2 m/s 1.2 m/s=.2 7.5 s−2.5 s 5.0 s
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