# B if the distance traveled in one direction is d then

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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 &lt; 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 x-intercept 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 x-intercept 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 x-intercept 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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