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Jifunze kwa makini maelezo haya kisha jibu maswali yafuatayo: "Maisha yetu ni mfano wa mbegu iliyopandwa. Mbegu nyingi hufa kabla hazijatoa mazao, lakini baadhi hukua na kutuletea furaha na mafanikio, na baadhi hukua na kutuletea huzuni na aibu. Hivyo basi, kila mmoja anapaswa kufanya maandalizi ya kutosha ili mbegu yake iweze kuota na kutoa mazao yanayompendeza."

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ScanSolve AI Answer

Step-by-step solution

26 steps

1(a)

  1. Convert speed to m/s.

    v=54×518=15 m/sv = 54 \times \frac{5}{18} = 15~\mathrm{m/s}
  2. Distance to cross the pole = length of train = 150 m.

  3. Time taken =distancespeed= \frac{\text{distance}}{\text{speed}}.

    t=150 m15 m/s=10 st = \frac{150~\mathrm{m}}{15~\mathrm{m/s}} = 10~\mathrm{s}

    10 s

    1(b)

  4. Convert speeds to m/s.

    Train 1: v1=54×518=15 m/sv_1 = 54 \times \frac{5}{18} = 15~\mathrm{m/s}

    Train 2: v2=36×518=10 m/sv_2 = 36 \times \frac{5}{18} = 10~\mathrm{m/s}

  5. Relative speed (opposite direction) =v1+v2=15+10=25 m/s= v_1 + v_2 = 15 + 10 = 25~\mathrm{m/s}.

  6. Distance to cross =150 m+200 m=350 m= 150~\mathrm{m} + 200~\mathrm{m} = 350~\mathrm{m}.

  7. Time taken =350 m25 m/s=14 s= \frac{350~\mathrm{m}}{25~\mathrm{m/s}} = 14~\mathrm{s}.

    14 s

    1(c)

  8. Convert speeds to m/s.

    Train 1: v1=15 m/sv_1 = 15~\mathrm{m/s}

    Train 2: v2=18×518=5 m/sv_2 = 18 \times \frac{5}{18} = 5~\mathrm{m/s}

  9. Relative speed (same direction) =v1−v2=15−5=10 m/s= v_1 - v_2 = 15 - 5 = 10~\mathrm{m/s}.

  10. Distance to cross =150 m+250 m=400 m= 150~\mathrm{m} + 250~\mathrm{m} = 400~\mathrm{m}.

  11. Time taken =400 m10 m/s=40 s= \frac{400~\mathrm{m}}{10~\mathrm{m/s}} = 40~\mathrm{s}.

    40 s

    2(a)

  12. Outward speed a=40 km/ha = 40~\mathrm{km/h}, return speed b=60 km/hb = 60~\mathrm{km/h}.

  13. Average speed for equal distances =2aba+b= \frac{2ab}{a + b}.

    2×40×6040+60=4800100=48 km/h\frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48~\mathrm{km/h}

    48 km/h

    2(b)

  14. Outward

    40 km at 50 km/h50~\mathrm{km/h}.

    Time =4050=0.8 h= \frac{40}{50} = 0.8~\mathrm{h}.

  15. Return

    60 km at 40 km/h40~\mathrm{km/h}.

    Time =6040=1.5 h= \frac{60}{40} = 1.5~\mathrm{h}.

  16. Total distance =40+60=100 km= 40 + 60 = 100~\mathrm{km}.

    Total time =0.8+1.5=2.3 h= 0.8 + 1.5 = 2.3~\mathrm{h}.

  17. Average speed =1002.3=1000023≈43.48 km/h= \frac{100}{2.3} = \frac{10000}{23} \approx 43.48~\mathrm{km/h}.

    \dfrac{10000}{23} km/h

    3(a)

  18. Initial speed u=72×518=20 m/su = 72 \times \frac{5}{18} = 20~\mathrm{m/s}.

    Final speed v=0 m/sv = 0~\mathrm{m/s}, a=−10 m/s2a = -10~\mathrm{m/s^2}.

  19. Stopping distance from v2=u2+2asv^2 = u^2 + 2as.

    0=(20)2+2(−10)s0 = (20)^2 + 2(-10)s 400=20s400 = 20s s=40020=20 ms = \frac{400}{20} = 20~\mathrm{m}

    20 m

    3(b)

  20. Initial speed u=0 m/su = 0~\mathrm{m/s}, a=2 m/s2a = 2~\mathrm{m/s^2}, t=10 st = 10~\mathrm{s}.

  21. Distance s=ut+12at2s = ut + \frac{1}{2}at^2.

    s=0+12×2×(10)2=100 ms = 0 + \frac{1}{2} \times 2 \times (10)^2 = 100~\mathrm{m}

    100 m

    3(c)

  22. Let length of train =L m= L~\mathrm{m}, speed =v m/s= v~\mathrm{m/s}.

    Time to pass man (point): Lv=18 s\frac{L}{v} = 18~\mathrm{s} ⇒L=18v\Rightarrow L = 18v.

  23. Time to pass 100 m100~\mathrm{m} platform

    L+100v=30 s\frac{L + 100}{v} = 30~\mathrm{s}.

  24. Subtract equations

    100v=12 s\frac{100}{v} = 12~\mathrm{s} ⇒v=10012 m/s\Rightarrow v = \frac{100}{12}~\mathrm{m/s}.

  25. L=18×10012=150 mL = 18 \times \frac{100}{12} = 150~\mathrm{m}.

    150 m, \dfrac{100}{12} m/s

    3(d)

  26. u=20 m/su = 20~\mathrm{m/s} upward, g=10 m/s2g = 10~\mathrm{m/s^2}.

    Max height h=u22gh = \frac{u^2}{2g}.

    h=(20)22×10=40020=20 mh = \frac{(20)^2}{2 \times 10} = \frac{400}{20} = 20~\mathrm{m}

    20 m

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