###### tags: `physics solution`
###### tags: `Motion along a straight line`
# Motion along a straight line
## Problem 1-9
1. ans:
> $90km/h=25m/s,d=25\times 0.5=12.5m$
2. ans:
> $(a)v=\frac{2\times 73.2}{\frac{73.2}{1.22}\ + \frac{73.2}{2.85}}=\frac{146.4}{85.684}=1.709m/s$
> $(b)v=\frac{1.22\times 60+3.05\times 60}{120}=2.135m/s$
> $(c)$略
3. ans:
> $(a)v=\frac{40+40}{\frac{40}{30}+\frac{40}{60}}=40km/h$
> $(b)v=average\ velocity=40km/h$
> $(c)$略
4. ans:
> Suppose that the single distance is $x$ km, $v=\frac{2x}{\frac{x}{35}+\frac{x}{60}}\simeq 44.211km/h$
5. ans:
> $(a)(b)(c)(d)$ $x=0、x=-2、x=0、x=12$
> $(e)\ d=+12$
> $(f)\ v=+7m/s$
> $(g)$略
6. ans:
> $19km/h\simeq 5.278m/s\rightarrow t=\frac{200}{\frac{200}{6.509}\ \ +19\times\frac{1000}{3600}}\simeq 5.554s$
7. ans:
> $d=\frac{60}{30+30}\times 60=60km$
8. ans:
> $(a)v\ or\ ratio=\frac{0.25}{\frac{1.75}{3.5}}=0.5m/s$
> $(b)t=\frac{5}{0.5}=10s$
9. ans:
> $\frac{L_2}{148.15}<\frac{L_1}{147.95}\rightarrow \frac{L_2}{L_1}<1.00136$ if $L_1=1km$, $L_2-L_1<0.00136km$

## Problem 10-26
10. ans:
> (a)method-1, suppose the wind-effects's velocity is $v_w$, car's velocity is $v_c$ $v_c+v_w=d/t_1$, $v_c-v_w=d/t_2$, so $v_c=\frac{\frac{d}{t_1}+\frac{d}{t_2}}{2}$
> (b)$\left\vert method_1-method_2\right\vert=\left\vert \frac{\frac{d}{t_1}+\frac{d}{t_2}}{2}-\frac{2d}{t_1+t_2}\right\vert=\left\vert v_c-\displaystyle\frac{2}{\frac{1}{v_c+v_w}+\frac{1}{v_c-v_w}}\right\vert=\left\vert v_c-\frac{v_c^2-v_w^2}{v_c}\right\vert=\left\vert \frac{v_w^2}{v_c} \right\vert\rightarrow$ Thus, percentage difference is $\left(\frac{v_w}{v_c}\right)^2\times 100\%=0.0576\%$
11. ans:
> $3h15min-2h=1h15min\rightarrow v=\frac{300-100-40}{75}=128km/h$
12. ans:
> (a)$\frac{d+L}{v}=\frac{L}{v_s}\rightarrow d=\frac{v}{v_s}L-L=48m$
> (b)Suppose the extra distance is $d_s$, $t=\frac{2d+d_s}{25}=\frac{d_s}{5}\rightarrow d_s=24m,t=4.8s\rightarrow$ the speed of shock wave is $\frac{24-12}{4.8}=2.5m/s$(subtract the fast car)
> $(c)$ because the wave's direction is in the traffic direction, it is $downstream$
13. ans:
> $(a)$Suppose time is $t$, $v=\displaystyle\frac{55\times\frac{1}{2}t+90\times\frac{1}{2}t}{t}=72.5km/h$
> $(b)$Suppose distance is $d$, $v=\displaystyle\frac{d}{\frac{0.5d}{55}+\frac{0.5d}{90}}\simeq 68.276km/h$
> $(c)d=27.5t+45t\rightarrow v=\displaystyle\frac{2d}{\frac{0.5d}{55}+\frac{0.5d}{90}+t}=\frac{145t}{\frac{36.25t}{55}+\frac{36.25t}{90}+t}\simeq 70.325km/h$
> $(c)$ another solution $v=\displaystyle\frac{2d}{\frac{d}{72.5}+\frac{d}{68.276}}\simeq 70.325km/h$
> $(d)\ 0$
> $(e)$略
14. ans:
> $v(t)=\frac{d}{dt}\left(16te^{-t}\right)=16\left(e^{-t}-te^{-t}\right),\ v=0,t=1\rightarrow x=\frac{16}{e}\simeq 5.886m$
15. ans:
> $(a)$ $v(t)=\frac{d}{dt}(18t+5t^2)=10t+18\rightarrow t=2.0,v=38$
> $(b)$ $v=\displaystyle\frac{x_{_{3.0s}}-x_{_{2.0s}}}{3.0-2.0}\ =43m/s$
16. ans:
> $x(t)=4.0-6.0t^2,\ v(t)=\frac{d}{dt}(4.0-6.0t^2)=-12.0t$
> $(a)$ $v=0,t=0$
> $(b)$ $v=0,x=4.0$
> $(c)(d)x=0,t=\pm\frac{2}{\sqrt{6}}$
> $(e)$ 略
> $(f)$ $+20t$
> $(g)$ increasing;
17. ans:
> $x(t)=9.75+1.5t^3,\ v(t)=4.5t^2$
> $(a)\ v=\frac{50.25-21.75}{3-2}=28.5cm/s$
> $(b)\ v=4.5\times 2^2=18cm/s$
> $(c)\ v=4.5\times 3^2=40.5cm/s$
> $(d)\ v=4.5\times (2.5)^2=28.125cm/s$
> $(e)\ x=\frac{50.25+21.75}{2}=36,t\simeq 2.604,v\simeq 30.514cm/s$
> $(f)$ 略
18. ans:
> $x(t)=12t^2-2t^3,\ v(t)=24t-6t^2,\ a(t)=24-12t$
> $(a)(b)(c)\ x=61.25m, v=10.5m/s,\ a=-18m/s^2$
> $(d)(e)\ v=0,\ t=0\ or\ 4\rightarrow t=0,x=0;t=4,x=192-128=64m(Maximum)$
> $(f)(g)\ a=0,t=2,v=24m/s$
> $(h)\ v=0,t=4\ or\ 0,t=4,a=-24$
> $(i)\ v=\frac{42-0}{3-0}=14m/s$
19. ans:
> $a=\frac{-30-18}{2.4}=-20m/s^2$
20. ans
> $x(t)=20t-5t^3\rightarrow v(t)=20-15t^2\rightarrow a(t)=-30t$
> $(a)(b)\ v=0,t=\frac{2}{\sqrt{3}};a=0,t=0$
> $(c)\ t\in\mathbb{R},(d)\ no$
21. ans
> (a) $v^2=v_0^2+2ad\rightarrow (0)^2=(36.1)^2+2\times 210a\rightarrow a\simeq -3.105m/s^2$
> (b) $v=v_0+at\rightarrow 0=36.1-3.105t\rightarrow t=11.626s$
22. ans
> $(a)\ m/s^2(b)\ m/s^3$
> $x(t)=4t^2-2t^3,v(t)=8t-6t^2,a(t)=8-12t$
> $(c)\ v=0,t=0\ or\ \frac{4}{3}\rightarrow t=0,x=0;t=\frac{4}{3},x=\frac{64}{27}=2.370m(Maximum)$
> $(d):d=\displaystyle\int^\frac{4}{3}_0 8t-6t^2dt-\int^4_\frac{4}{3}8t-6t^2dt=68.741m;\\(e):\displaystyle\int^4_0 8t-6t^2=-64m$
> $(f)~(i)\ +2m/s,-8m/s,-30m/s,-64m/s$
> $(j)~(m)\ -4m/s^2,-16m/s^2,-28m/s^2,-40m/s^2$
23. ans
> $\displaystyle\frac{\frac{1}{2}a5^2-\frac{1}{2}a4^2}{\frac{1}{2}a5^2}=\frac{4.5}{12.5}=0.36$
24. ans
> $(a)\ (1.6)^2=2a\times 5\times 10^{-5},a=2.56\times 10^4m/s^2\simeq 0.261\times 10^4g$
> $(b)\ 0=(1.6)^2+2a\times 10^{-3},a=-1.28\times 10^3m/s^2\simeq -0.131\times 10^3g$
25. ans
> $(2.8)^2=v_0^2+2a\times 6,2.8=v_0+2.5a,6=2.5v_0+\frac{1}{2}a(2.5)^2\rightarrow a=-0.32m/s^2$
26. ans
> (a) $0=(6\times 10^6)^2+2\times 1.25\times10^{14}S,S=-0.144m$
> (b) 略

## Problem 27-43
27. ans
> (a) $v=9.6-2.5\times 3.2=1.6m/s$
> (b) $v=9.6+2.5\times 3.2=17.6m/s$
28. ans
> (a) $\frac{27.2}{4.92}=5.528m/s^2$
> (b) $d=\frac{1}{2}4.92\times \left(\frac{27.2}{4.92}\right)^2=75.187m$
> $(c)$ 略
29. ans
> (a) $(\frac{305}{60})^2=0+2\times 1.22S\rightarrow S\simeq10.59m$
> (b) $t=\frac{2\times 305}{60\times 1.22}+\frac{190-2\times 10.59}{5.083}\simeq 41.546s$
30. ans
> (a) $t=\frac{90-146}{-18.72}=2.991s$
> (b) 略
31. ans
> $S=vt+\frac{1}{2}at^2,\rightarrow t=10s,v=100\rightarrow S_{max}=500+100\times (\frac{100}{9.8})+\frac{1}{2}(-9.8)\left(\frac{100}{9.8}\right)^2$
> $=500+\frac{1}{2}\times 9.8\times\left(\frac{100}{9.8}\right)^2=1010.204m$
32. ans
> $a=-\displaystyle\frac{1020\times 1000/3600}{1.4}=-202.381m/s^2=-20.651g$
33. ans
> (a) $56km/h\simeq 15.556m/s,\ 24=15.556\times 2+\frac{1}{2}a\times 2^2,\ a\simeq -3.556m/s^2$
> (b) $v=15.\overline{5}-2\times 3.\overline{5}=8.\overline{4}\simeq 8.4m/s$
34. ans
> $S=v_0t+\frac{1}{2}at^2,t=\frac{S}{v}$
> $\rightarrow situation_1,t=\frac{44.5}{20\times 1000/3600},175.5=v_0t+\frac{1}{2}at^2$
> $\rightarrow situation_2,t=\frac{77.9}{40\times 1000/3600},142.1=v_0t+\frac{1}{2}at^2$
> $v_0=-8.745m/s,a=-3.287m/s^2$
35. ans???
> $v^2=v_0^2+2aS\rightarrow a=\frac{(6\times 10^7)^2-(4\times 10 ^5)^2}{2\times 3\times 10^{-2}},t=\frac{\Delta v}{a}=9.934 \times 10^{-9}s$
36. ans(題目數值有誤)
> (a) $\frac{1}{2}\times 2.75t_1^2=225,\ t_1=12.792s;\ 675=\frac{1}{2}\times 0.75 t_2^2,t_2=42.426s,\ t=t_1+t_2=55.218s$
> (b) $v=at=2.75\times 12.792=35.178m/s$
> $(c)$ 略
37. ans
> $a=constant\rightarrow x(t)=at^2+bt+c,\ a,b,c\in\mathbb{R}$
> $\rightarrow \left\{ \begin{array}{} a+b+c=0 \\ 4a+2b+c=6 \\ c=-2\end{array} \right. \rightarrow x(t)=2t^2-2,\ a=4m/s^2,right(\rightarrow)$
38. ans
> $(a)$ $\frac{880}{2}=\frac{1}{2}\times 1.34t^2,\ t=25.626s,\ v=34.339m/s$
> $(b)$ $t=2\times 25.626=51.252s$
> $(c)$ $v=\frac{880}{20+51.252}=12.351m/s$
> $(d)$ 略
39. ans
> $46(t+1)=\frac{1}{2}\times 4 t^2,\ t=23.96s$
40. ans
> $55km/h=15.278m/s,\ 0.75\times 55\times\frac{1000}{3600}=11.458m,\rightarrow\\ \frac{1}{2}\times5.18t^2=40-11.458,\ t=3.32> 2.8\rightarrow inappropriate$
> $\frac{1}{2}\times 5.18t^2=32-11.458,\ t=2.816> 1.8\rightarrow inappropriate$
> So you should brake to a stop
41. ans
> (a) $\frac{64}{3}m/s^2$
> (b) $\frac{1}{2}\times\frac{64}{3}\times3^2=96m$
42. ans
> $120km/h=33.\overline{3}m/s$
> $(a)\ d=25-\left(\ 33.333\times 2 -\left(33.333\times 2-\frac{1}{2}\times 5 \times 2^2\right)\right)=15m$
> $(b)\ d=25-\left(33.333\times 2.4-\left(33.333\times 2.4-\frac{1}{2}\times 5\times (2.4)^2\right)\right)=10.6m$
> $\rightarrow 10.6=5\times 2.4t,\ t=0.883s,\ v=33.333-5\times 0.883=28.916m/s$
43. ans
> (a) $v(compared)=36.667m/s,\ 676=36.667t-\frac{1}{2}at^2,36.667=at$
> $\\ \rightarrow t=\displaystyle\frac{676}{36.667-0.5\times36.667}\rightarrow a=\frac{v}{t}=\frac{0.5(36.667)^2}{676}=0.994m/s^2$
> (b) 略

## Problem 44-65
44. ans
> $(a)\ 0.558=v_0\times0.2-\frac{1}{2}\times 9.8\times (0.2)^2\rightarrow v_0=3.92m/s$
> $(b)\ v^2=\left(3.92\right)^2-2\times 9.80\times 0.544\rightarrow v=2.169m/s$
> $(c)\ h=\frac{1}{2}\times 9.8\times \left(\frac{3.92}{9.8}\right)^2=0.784m$
> $(c)\ another\ solution:h=\frac{v^2_f-v^2_i}{2a}=\frac{(3.92)^2}{2\times9.8}=0.784m$
45. ans
> $(a)$ $50=\frac{1}{2}\times 9.8\times t^2,\ t=3.194,v=31.305m/s$
> $(b)$ $t=2\times 3.194=6.388s$
> $(c)$ 略
46. ans
> (a) $v=\sqrt{2\times 9.8\times 1800}=-187.83m/s$
> (b) safe
47. ans
> $(a)\ h=\frac{(24)^2}{2\times 9.8}=29.388m$
> $(b)\ t=\sqrt{\frac{2h}{a}}=2.449s$
> $(c)\ 略$
48. ans
> (a) $t=\frac{28.513-15}{9.8}=1.379s$
> (b) $v=\sqrt{(15)^2+2\times9.8\times 30}=-28.513m/s$
49. ans
> (a) $h_{up}=\frac{(12)^2}{2\times 9.8},h_{up}=7.347m,t_1=\frac{12}{9.8}=1.224s$
> $\rightarrow 80+7.347=\frac{1}{2}\times 9.8t_2^2,t_2=4.222s,t=t_1+t_2=5.447s$
> (b) $v=9.8\times -4.222=-41.3756m/s$
50. ans
> (a) $29.4m/s$
> (b) ??? $3\times 9.8=29.4m/s$
51. ans
> $v=\sqrt{2\times 9.8\times \frac{39.2}{2}}=19.6m/s$
52. ans
> $(a)$ $80=\frac{1}{2}\times 9.8t^2,\ t=4.041s$
> $(b)$ $v=9.8\times 4.041=39.598m/s$
> $(c)$ $v=\sqrt{2\times 9.8 \times 100}=44.272m/s$
53. ans
> $v=\displaystyle\frac{12}{\sqrt{\frac{2\times 45}{9.8}}}=3.96m/s$
54. ans
> $stone_1:t=\sqrt{\frac{2\times 53.6}{9.8}},\ stone_2:53.6=v(t-1)+\frac{1}{2}\times 9.8\times (t-1)^2\rightarrow v=11.929m/s$
55. ans
> (a) $v=\sqrt{2\times 9.8\times 15}=17.146m/s\rightarrow a=\frac{v}{t}=\frac{17.146}{20\times 10^{-3}}=8.573\times 10^2m/s^2$
> (b) up
56. ans
> $\left(\frac{1}{3}v_A\right)^2=v^2_A-2\times 9.8\times 0.4\rightarrow v_A=8.82m/s$
57. ans
> (a) $a=\displaystyle\frac{\sqrt{2\times 9.8\times 2}+\sqrt{2\times 9.8\times 4}}{12\times 10^{-3}}=1.26\times 10^3m/s^2$
> (b) up
58. ans
> $\frac{1}{2}gt^2=h,\ \frac{1}{2}g(t-1)^2=0.4h,\ 0.6h=g(t-1)\times 1+\frac{1}{2}g1^2$
> $\rightarrow 0.3t^2=(t-1)+0.5\rightarrow t=2.721s\ or\ 0.613s(0.613-1<0),h=36.279m$
59. ans
> (a) $t=\sqrt{\frac{2\times 2}{9.8}}=0.639\rightarrow h_2=2-\frac{1}{2}\times 9.8 \left(\frac{2}{3}\sqrt{\frac{4}{9.8}}\right)^2=0.889m$
> (b) $h_3=2-\frac{1}{2}\times 9.8\left(\frac{1}{3}\sqrt{\frac{4}{9.8}}\right)^2=0.222m$
60. ans
> $h=v_0t+\frac{1}{2}at^2=9.8\times 2.5\times 1.5 -\frac{1}{2}\times 9.8\times (1.5)^2=25.725m$
61. ans
> Suppose the final(max) velocity is $v_f$, $h=\frac{1}{2}\times gt^2=\frac{1}{2}\times\frac{(v_f)^2}{g}$
> $\left(v_f-g\right)^2=\left(v_f-1.125g\right)^2+2g\times 1.2\rightarrow v_f=20.0125m/s\rightarrow h=20.437m$
62. ans
> (a) $v=\sqrt{2\times 9.8\times 0.15}=1.715m/s,\ t=\displaystyle\frac{2v}{g}=0.35s$
> (b) $v=\sqrt{2\times 9.8\times 0.63}=3.514m/s,\ t=\frac{2v}{g}=0.717s$
63. ans
> Suppose that when flowerpot pass the top of window,velocity is $v$, $h=\frac{v^2}{2g},\ 2=0.25v+\frac{1}{2}\times 9.8\times (0.25)^2$
> $v=6.775m/s,\ h=2.342m$
64. ans
> $h=10\times 3+\frac{1}{2}\times 9.8\times 3^2=74.1m$
65. ans
> (a) $v=0.5\times 50\times 10^{-3}\times 90=2.25m/s$
> (b) $v=(80+20)\times 50\times 0.5+(50+20)\times 40\times 0.5=3.9m/s$

## Problem 66-70
66. ans
> (a) $d=10\times 10^{-3}\times 2\times 0.5+(2+4)\times 40\times 10^{-3}\times 0.5=0.13m$
> (b) $0.12s$
67. ans
> $\left\vert \left\{120\times 2\times 0.5+\left(120+140\right)\times 2\times 0.5+(140+200)\times 2\times 0.5+200\times 0.5\right\}-\\ \left\{ 40\times 3\times 0.5+40+(40+80)\times 2\times 0.5+80\times 0.5\right\} \right\vert\times 10^{-3}=0.56m/s$
68. ans
> $v=(10\times 100\times 0.5+(100+400)\times 10 \times0.5+400\times 10\times 0.5)\times 10^{-3}=5m/s$
69. ans
> $d=8\times 2\times 0.5+8\times 8+(8+4)\times 2\times 0.5+4\times 4=100m$
70. ans
> $p_1's\ v(t)=12t+3;\ p_2's\ v(t)=-4t^2+15\rightarrow 12t+3=-4t^2+15$
> $t=\displaystyle\frac{-3+\sqrt{21}}{2}, v=12.495m/s$