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."

1(a)

Step 1: Convert speed to m/s.

$v = 54 \times \frac{5}{18} = 15~\mathrm{m/s}$

Step 2: Distance to cross the pole = length of train = $150$ m.

Step 3: Time taken $= \frac{distance}{speed}$.

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

10 s

1(b)

Step 1: Convert speeds to m/s.

Train 1: $v_1 = 54 \times \frac{5}{18} = 15~\mathrm{m/s}$

Train 2: $v_2 = 36 \times \frac{5}{18} = 10~\mathrm{m/s}$

Step 2: Relative speed (opposite direction) $= v_1 + v_2 = 15 + 10 = 25~\mathrm{m/s}$.

Step 3: Distance to cross $= 150~m + 200~m = 350~\mathrm{m}$.

Step 4: Time taken $= \frac{350~m}{25~m/s} = 14~\mathrm{s}$.

14 s

1(c)

Step 1: Convert speeds to m/s.

Train 1: $v_1 = 15~\mathrm{m/s}$

Train 2: $v_2 = 18 \times \frac{5}{18} = 5~\mathrm{m/s}$

Step 2: Relative speed (same direction) $= v_1 - v_2 = 15 - 5 = 10~\mathrm{m/s}$.

Step 3: Distance to cross $= 150~m + 250~m = 400~\mathrm{m}$.

Step 4: Time taken $= \frac{400~m}{10~m/s} = 40~\mathrm{s}$.

40 s

2(a)

Step 1: Outward speed $a = 40~\mathrm{km/h}$, return speed $b = 60~\mathrm{km/h}$.

Step 2: Average speed for equal distances $= \frac{2ab}{a + b}$.

$\frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48~km/h$

48 km/h

2(b)

Step 1: Outward: $40$ km at $50~\mathrm{km/h}$.

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

Step 2: Return: $60$ km at $40~\mathrm{km/h}$.

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

Step 3: Total distance $= 40 + 60 = 100~\mathrm{km}$.

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

Step 4: Average speed $= \frac{100}{2.3} = \frac{10000}{23} \approx 43.48~\mathrm{km/h}$.

\dfrac{10000{23} km/h}

3(a)

Step 1: Initial speed $u = 72 \times \frac{5}{18} = 20~\mathrm{m/s}$.

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

Step 2: Stopping distance from $v^2 = u^2 + 2as$.

$0 = (20)^2 + 2(-10)s$

$400 = 20s$

$s = \frac{400}{20} = 20~\mathrm{m}$

20 m

3(b)

Step 1: Initial speed $u = 0~\mathrm{m/s}$, $a = 2~\mathrm{m/s^2}$, $t = 10~\mathrm{s}$.

Step 2: Distance $s = ut + \frac{1}{2}at^2$.

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

100 m

3(c)

Step 1: Let length of train $= L~\mathrm{m}$, speed $= v~\mathrm{m/s}$.

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

Step 2: Time to pass $100~\mathrm{m}$ platform: $\frac{L + 100}{v} = 30~\mathrm{s}$.

Step 3: Subtract equations: $\frac{100}{v} = 12~\mathrm{s}$ $\Rightarrow v = \frac{100}{12}~\mathrm{m/s}$.

Step 4: $L = 18 \times \frac{100}{12} = 150~\mathrm{m}$.

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

3(d)

Step 1: $u = 20~\mathrm{m/s}$ upward, $g = 10~\mathrm{m/s^2}$.

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

$h = \frac{(20)^2}{2 \times 10} = \frac{400}{20} = 20~\mathrm{m}$

20 m