Link download full: Solutions Manual Physics for Scientists and Engineers 3rd by Fishbane https://testbankservice.com/download/solutions- manual-fishbane-physics-for-scientists-and- engineers-3rd/ CHAPTER 2 Straight-Line Motion Answers to Understanding the Concepts Questions 1. You should be worried about something that might happen to bring the car in front of you to a stop. Your stopping time depends on two factors: Your fixed reaction time, and the time required for your brakes to bring your car to a stop. At a higher initial speed, you travel farther during the time it takes you to react and apply the brakes, and you travel farther in the time it takes your brakes to bring you to a stop. Both factors, then, argue for increasing spacing with increasing speed. 2. The velocity of the chalk is zero at that point, but the acceleration remains g, downward. In fact if the acceleration were zero as well then the chalk would maintain zero velocity there –– i.e., it would “freeze” at the top of its path! 3. We have seen that for a fixed acceleration gx, the relation between fall distance d and fall time t is, assuming the falling object starts from rest, d = !gxt 2 . Thus for fixed d, t = (2d/gx) 1/2 . The variation from planet to planet for t, that is, with gx, is then (gx) – l/2 . The larger gx, the smaller the fall time. The speed of the object at the end of the fall is v = gxt = (2dgx) 1/2 . The speed increases with gx like (gx) 1/2 . 4. Both you and the bowling balls would be falling at the same rate (g), so there is no reason to worry that any of them would crash onto you. 5. The acceleration of a falling object is equal to g only in a true free fall, which is devoid of any air resistance. In reality, as the object falls, it encounters an air resistance which increases with its speed. Initially, the object is not moving very fast so the air resistance exerted on it is not yet significant, and it falls with an acceleration close to g. As it speeds up, however, it picks up more and more air resistance so its acceleration gradually diminishes, until it reaches a certain speed, at which the upward air resistance equals the downward gravitational pull, whereupon its acceleration is zero and the its speed can no longer increase. This speed is therefore referred to as the terminal speed. So no, the speed of a falling object cannot increase indefinitely. 6. Certainly if the (negative) acceleration has a constant magnitude, the velocity cannot remain positive. Indeed, if the initial velocity has magnitude v0, and the acceleration has the constant magnitude a, then the velocity varies with time according to v = v0 – at, and v = 0 at a time t = v0/a; for times greater than this the velocity is negative. However, the acceleration could steadily decrease in magnitude, while remaining negative, such that the velocity could remain positive. A physical example occurs when a rocket is sent away from Earth with enough initial speed to leave the Solar System--we say that its initial speed exceeds the “escape speed.” If we say that “up” is the positive direction, then the acceleration is negative while the velocity is positive. As the object moves away, the force of gravity on it, and hence its acceleration, decreases in magnitude while remaining negative. For a fast enough start, the object never comes to rest or turns around. Incidentally, the escape speed from Earth is about 11.2 km/s.
34
Embed
Solutions Manual Physics for Scientists and Engineers 3rd · manual-fishbane-physics-for-scientists-and-engineers-3rd/ CHAPTER 2 Straight-Line Motion Answers to Understanding the
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Link download full: Solutions Manual Physics for
Scientists and Engineers 3rd by Fishbane
https://testbankservice.com/download/solutions-
manual-fishbane-physics-for-scientists-and-
engineers-3rd/
CHAPTER 2 Straight-Line Motion
Answers to Understanding the Concepts Questions 1. You should be worried about something that might happen to bring the car in front of you to a stop. Your
stopping time depends on two factors: Your fixed reaction time, and the time required for your brakes to
bring your car to a stop. At a higher initial speed, you travel farther during the time it takes you to react
and apply the brakes, and you travel farther in the time it takes your brakes to bring you to a stop. Both
factors, then, argue for increasing spacing with increasing speed.
2. The velocity of the chalk is zero at that point, but the acceleration remains g, downward. In fact if the
acceleration were zero as well then the chalk would maintain zero velocity there –– i.e., it would “freeze”
at the top of its path!
3. We have seen that for a fixed acceleration gx, the relation between fall distance d and fall time t is,
assuming the falling object starts from rest, d = !gxt2. Thus for fixed d, t = (2d/gx)1/2. The variation from
planet to planet for t, that is, with gx, is then (gx) – l/2. The larger gx, the smaller the fall time. The speed of
the object at the end of the fall is v = gxt = (2dgx)1/2. The speed increases with gx like (gx)1/2.
4. Both you and the bowling balls would be falling at the same rate (g), so there is no reason to worry that
any of them would crash onto you.
5. The acceleration of a falling object is equal to g only in a true free fall, which is devoid of any air
resistance. In reality, as the object falls, it encounters an air resistance which increases with its speed.
Initially, the object is not moving very fast so the air resistance exerted on it is not yet significant, and it
falls with an acceleration close to g. As it speeds up, however, it picks up more and more air resistance
so its acceleration gradually diminishes, until it reaches a certain speed, at which the upward air
resistance equals the downward gravitational pull, whereupon its acceleration is zero and the its speed
can no longer increase. This speed is therefore referred to as the terminal speed. So no, the speed of a
falling object cannot increase indefinitely.
6. Certainly if the (negative) acceleration has a constant magnitude, the velocity cannot remain positive.
Indeed, if the initial velocity has magnitude v0, and the acceleration has the constant magnitude a, then
the velocity varies with time according to v = v0 – at, and v = 0 at a time t = v0/a; for times greater than
this the velocity is negative. However, the acceleration could steadily decrease in magnitude, while
remaining negative, such that the velocity could remain positive. A physical example occurs when a
rocket is sent away from Earth with enough initial speed to leave the Solar System--we say that its initial
speed exceeds the “escape speed.” If we say that “up” is the positive direction, then the acceleration is
negative while the velocity is positive. As the object moves away, the force of gravity on it, and hence its
acceleration, decreases in magnitude while remaining negative. For a fast enough start, the object never
comes to rest or turns around. Incidentally, the escape speed from Earth is about 11.2 km/s.
25. The particle never gets farther from the origin than A; it oscillates back and forth through the origin. The magnitude of the velocity is maximum at the origin and zero at x = ±A.
The magnitude of the acceleration is maximum at x = ±A and zero at the origin.
2 at at
26. From 2 atx = At + Be we can get the speed: v = dx/dt = 2At + Bae and the acceleration: a = dv/dt = 2A +
B a e .
2 a(0) Putting the initial conditions into x, we get – 1.5 m = A(0) + Be , or B = – 1.5 m. a(0)
Putting the initial conditions into v, we get 0.25 m/s = 2A(0) + Bae ;
-1
Ba = 0.25 m/s and thus a = – 0.17 s .
–0.1/6,
To obtain A, we use the second velocity condition: 0.045 m/s = 2A(0.10 s) + (– 0.25 m/s)e 2
which gives A = –1.0 m/s . At t = 1.0 s the acceleration is
2 -1 2 (–1/6) 2 a = 2(–1.0 m/s ) + (– 1.5
m/s)(–0.17 s ) e = – 2.0 m /s .
27. We see from a = A – v/t0 that the acceleration is a function of velocity and thus time.
At t = 0, a = A – v0/t0. If this is positive, the velocity will increase, which causes a decrease in the
acceleration. The rate of change of the velocity (the slope of the v-t curve) will decrease. Eventually
the acceleration becomes zero, at which point the velocity becomes constant (called a terminal
velocity). After a long time a = 0, so A – vterminal/t0 = 0, or vterminal = At0 .
The graphs assume that A > v0/t0.
2 2 2
28. For constant acceleration we have x = x0 + v0t + !at and v = v0 + 2a(x – x0).
2 4 2
A – ( v 0 / t 0 )
O
Time
A t 0
v 0
O Time
Chapter 2: Straight-Line Motion
From the v-equation: (128 mi/h) = 0 + 2a[# mi/ – 0], which gives a = 3.28 ⋅ 10 mi/h . From
4 2 2
the x-equation: # mi = 0 + 0 + ! (3.28 ⋅ 10 mi/h )t ,
which gives t = 0.00391 h = 14.1 s.
3
29. We convert the speeds to ft/s: (25 mi/h)(5.28 ⋅ 10 ft/mi)/(3600 s/h) = 36.7 ft/s; 50 mi/h = 73.3 ft/s. For
. The height reached is determined by the initial velocity. We assume the same initial velocity of the
2 2
jumper on the Moon and Earth. With a vertical velocity of 0 at the highest point, from v = v0 + 2ah we
get 2
v0 = 2gEh E = 2gMh M, or
h M = (gE/gM)h E = (6)(2 m) = 12 m.
74. The speed attained in a fall through a height h can be found from
2 2 2 v1 = v0 +
2a(y – y0); v1 = 0 + 2gh.
If we assume you stop by flexing your knees a distance 2 2
∆: v2 = v1 + 2a(y – y0); 0 = 2gh + 2a∆, which 2
gives a = – g h/∆ = – 9.8(8 ft/∆ ft) m/s . 75. We use a coordinate system with the origin at the throw and up positive. Then v0 = 7 m/s. From v = v0
+ at = v0 – gt;
– 4 m/s = 7 m/s – g(7 s), we get g 2
m/s . To find the highest point, we have 2 2
v = v0 + 2a(y – y0); 2 2
0 = (7 m/s) + 2(– 1.6 m/s )h, which gives h = 15 m . 76. We use a coordinate system with the origin at June’s initial position and the positive direction toward Bill. 2 2
Then xaJ = 1.0 m/s and 2 a= 0 + 0 + !(1.0 m/sB = – 0.9 m/s2)t:2 .
73
= 1.6
Page 2-21
x BJ == x
x00JB
+0+ 0 v t v t + + ! a t!aJ B t 2 = 20 m + 0 + 2!(– 0.9 m/s2)t2. 2
When they meet: x = x , or 0.50t = 20 – 0.45t , which gives
J B t = 4.6 s and xJ = 11 m .
77 . Let da/dt = J = constant. Integrate over time: a(t) = a0 + Jt, where a0 is the initial
78. v(t) = dx/dt = 4t3 – t, so the displacement from t1 to t2 is
∆
x = ∫t t12 v(t)dt =∫t t12 (4t 3 − t)dt = t 4 − 12 t 2 tt 21 2
2 .
Plug in t1 = 0.5 s and t2 = 1.5 s to obtain ∆x = 4 m.
The average speed is vav = ∆x/∆t =3 4
m/(1.5 s – 20.5 s) = 4 m/ s.
a(t) = dv/dt = d(4t – t)/dt = 12t – 1, which is negative before t1 = √(1/12) s = 0.29 s and positive
thereafter; while v(t) = 4t3 – t, which is negative before t2 = 0.5 s and positive thereafter. So between 0.5 s
and 1.5 s both v and a have the same direction, indicating that the speed increases monotonically during
this time interval. So vmin occurs at t = 0.5 s while vmax at t = 1.5 s. Plug these values of t into the
expression v(t) = 4t3 – t to obtain vmin = 0 and vmax = 12 m/s .
Note that vav = 4 m/s is indeed within this range. [Since the acceleration is not a constant here,
however, we generally cannot write vav = !(vmin + vmax).] 79. (a) Take up as the y-axis. The velocity of the ball just before it hits the switch after it falls through a
distance h is
v ri = (2gh) 1/2 (– j )=ˆ [2(9.8 m/s )(10
ˆ m )]2 1/2
.
(– j ) = (b) The ball rebounds to a height h’, so its the velocity as it leaves the switch is
(−14 m/s )j ˆ
4 1 2 4 1 2
= t 2 − t 2 − t 1 − t 1
Page 2-22
v
rf = (2gh’)
1/
2 ˆj = [2(9.8 m/s )(9 m )] j 2=
1
/2 ˆ
.
(c) ra av = (v f rr– v i)/∆t = [(13.3 m/s ) j ˆ –
(–14 m/s) j ˆ ]/0.002 s = (1.4⋅ 10 4 m/s 2 )jˆ .
80. Our coordinate system has the origin at the archer, up positive and t = 0 when the balloon is dropped.
For the balloon: 2
2yB2 = y0B + v0B t + !at = 200 m + 0 + ! (– 9.8
m/s )t . For the arrow:
2 2
yA = y0A + v0A t + !gt = 0 + (40 m/s)(t – 5 s) + !(– 9.8 m/s )(t – 5 2
s) . The arrow intercepts the 2 balloon when yB = yA:
2 2
2
(200 m) – !(9.8 m/s )t = (40 m/s)(t – 5) – !(9.8 m/s )(t – 5 s) . _ 2 2
The time of intercept is t = 5.87 s, which means y = 200 m – !(9.8 m/s )(5.87 s) = 31 m.
81. We use a coordinate system with the origin at the top of the snowbank with down positive. To 2 2
find the distance: 2v = v0 + 2a(y – y20);
0 = (40 m/s) + 2(– 50)(9.8 m/s )(y – y0), which gives y – y0 = 1.6
m. To find the time: v = v0 + at; 2
0 = 40 m/s + (– 50)(9.8 m/s )t, which gives t = 0.08 s .
82. We use a coordinate system with the origin at the ground and up positive. For all the weights, v0 = 0 and
a = – g. Take t = 0 when the string is released, and the lowest weight is at the ground. The time for the