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Unformatted text preview: 29. (c) x = −3 (a) The acceleration is the slope of the velocity, so a =
4
4
13
(b) v − 3 = − (t − 1), or v = − t +
3
3
3 30. 4
3 − (−1)
= − ft/s2 .
1−4
3
13
(c) v =
ft/s
3
5
−1
0−5
=−
=
ft/s2 .
10 − 0
10
2
(d) t = 4 s The acceleration is the slope of the velocity, so a = (b) v = 5 ft/s (a) It moves (to the left) 6 units with velocity v = −3 cm/s, then remains motionless for 5 s, then
moves 3 units to the left with velocity v = −1 cm/s. (b) vave = (c) 31. (a) Since the motion is in one direction only, the speed is the negative of the velocity, so
9
save =
cm/s.
10 (c) v = 4 ft/s 9
0−9
=−
cm/s
10 − 0
10 32. It moves right with constant velocity v = 5 km/h; then accelerates; then moves with constant, though
increased, velocity again; then slows down. 33. (a) If x1 denotes the ﬁnal position and x0 the initial position, then v = (x1 − x0 )/(t1 − t0 ) = 0 mi/h,
since x1 = x0 . (b) If the distance traveled in one direction is d, then the outward journey took t = d/40 h. Thus
2d
80t
total dist
=
=
= 48 mi/h.
save =
total time
t + (2/3)t
t + (2/3)t
(c)
34. 35. t + (2/3)t = 5, so t = 3 and 2d = 80t = 240 mi round trip (a) down, since v < 0 (a) (b) v
100
80
60
40
20
20 40 60 80 100 120 36. (c) It’s constant at 32 ft/s2 . (b) v = 0 at t = 2 x t t if
0 ≤ t ≤ 10 10t
100
if 10 ≤ t ≤ 100
v= 600 − 5t if 100 ≤ t ≤ 120 23 37. Chapter 1 (a) y = 20 − 15 = 5 when x = 45, so 5 = 45k ,
k = 1/9, y = x/9 (b) y
0.6
0.4
0.2 2 4 6 x (c) 38. l = 15 + y = 15 + 100(1/9) = 26.11 in. (d) If ymax = 15 then solve 15 = kx = x/9 for
x = 135 lb. (a) Since y = 0.2 = (1)k , k = 1/5 and y = x/5 (b) y
1 2 (c) y = 3k = 3/5 so 0.6 ft. (d) 4 6 x ymax = (1/2)3 = 1.5 so solve 1.5 = x/5 for
x = 7.5 tons 39. Each increment of 1 in the value of x yields the increment of 1.2 for y , so the relationship is linear. If
y = mx + b then m = 1.2; from x = 0, y = 2, follows b = 2, so y = 1.2x + 2 40. Each increment of 1 in the value of x yields the increment of −2.1 for y , so the relationship is linear.
If y = mx + b then m = −2.1; from x = 0, y = 10.5 follows b = 10.5, so y = −2.1x + 10.5 41. (a) With TF as independent variable, we have
(b) 5/9 (c) Set TF = TC = (d) 37◦ C TC − 100
5
0 − 100
=
, so TC = (TF − 32).
TF − 212
32 − 212
9 5
(TF − 32) and solve for TF : TF = TC = −40◦ (F or C).
9 (a) One degree Celsius is one degree Kelvin, so the slope is the ratio 1/1 = 1. Thus TC = TK − 273.15. (b) TC = 0 − 273.15 = −273.15◦ C 43. (a) 5.9 − 1
p−1
=
, or p = 0.098h + 1
h−0
50 − 0 44. (a) 42. (b)
45. (a)
(b) 46. (b) when p = 2, or h = 1/0.098 ≈ 10.20 m 133.9 − 123.4
R − 123.4
=
, so R = 0.42T + 115.
T − 20
45 − 20
T = 32.38◦ C
0.75 − 0.80
r − 0.80
=
, so r = −0.0125t + 0.8
t−0
4−0
64 days (a) Let the position at rest be y0 . Then y0 + y = y0 + kx; with x = 11 we get y0 + kx = y0 + 11k = 40,
and with x = 24 we get y0 + kx = y0 + 24k = 60. Solve to get k = 20/13 and y0 = 300/13. (b) 300/13 + (20/13)W = 30, so W = (390 − 300)/20 = 9/2 g. Exercise Set 1.6 47. (a) 24 For x trips we have C1 = 2x and
C2 = 25 + x/4 (b) 2x = 25 + x/4, or x = 100/7, so the commuter pass becomes worthwhile at x = 15. C
60 40 20 x
5 10 15 20 25 30 48. If the student drives x miles, then the total costs would be CA = 4000 + (1.25/20)x and
CB = 5500 + (1.25/30)x. Set 4000 + 5x/80 = 5500 + 5x/120 and solve for x = 72, 000 mi. 49. (a) H ≈ 20000/110 ≈ 181 (b) One light year is 9.408 × 1012 km and t = 1
1
9.408 × 1018 km
d
=
=
=
v
H
20km/s/Mly
20km/s = 4.704 × 1017 s = 1.492 × 1010 years.
(c) The Universe would be even older. EXERCISE SET 1.6
1. (a)
(c) y = 3x + b (b) y y = 3x + 6 10 y = 3x + 2 5
2 y = 3x + 6 y = 3x  4 1 1 2 x 5
10 2.
3. 1
Since the slopes are negative reciprocals, y = − x + b.
3
(a)
(c) (b) y = mx + 2 m = tan φ = tan 135◦ = −1, so y = −x + 2 (b) y = m(x − 1) (d) 2x + 4y = C y
m = 1.5 5
m = –1 4
m=1 3
2
1
2 1 1 2 x 1 4. (a) y = mx (c) y = −2 + m(x − 1) 25 5. Chapter 1 The slope is −1. (a) (b) The y intercept is y = −1.
y y
4 5
4
3
2
1
2 (c) 1 2
2
1 1
2
3 1 1
2 x 2 x 2 4
6 They pass through the point (−4, 2). (d) The xintercept is x = 1.
y y
3
6 2 4 1
x 2 0.5 4 2 1.5 1 x
6 1 2 2 2 3 6. (a) horizontal lines (b) The y intercept is y = −1/2. y y
2 x x
2
1 (c) The xintercept is x = −1/2. (d) They pass through (−1, 1). y y (1,1)
1
1
x
1 x
2 1 7. 2
Let the line be tangent to the circle at the point (x0 , y0 ) where x2 + y0 = 9. The slope of the tangent
0
line is the negative reciprocal of y0 /x0 (why?), so m = −x0 /y0 and y = −(x0 /y0 )x + b. Substituting
9 − x0 x
.
the point (x0 , y0 ) as well as y0 = ± 9 − x2 we get y = ±
0
9 − x2
0 8. Solve the simultaneous equations to get the point (−2, 1/3) of intersection. Then y = 1
+ m(x + 2).
3 Exercise Set 1.6 9. 26 The xintercept is x = 10 so that with depreciation at 10% per year the ﬁnal value is always zero, and
hence y = m(x − 10). The y ...
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