F hours after the departure of the bus from P. (a) Express the average speed of the trailer in terms of t. (b) Find the ratio of the speed of the bus to that of the trailer.

Tunajua umbali $D = 3120$ km.
Acha $S_B$ iwe kasi ya lori B (km/h).
Acha $S_A$ iwe kasi ya lori A (km/h).
Acha $T_B$ iwe muda uliotumika na lori B (masaa).
Acha $T_A$ iwe muda uliotumika na lori A (masaa).

Kutoka kwa swali, tunajua:

• Kasi ya lori A ni 5 km/h haraka kuliko lori B: $S_A = S_B + 5$.
• Lori B inachukua masaa 4 zaidi kuliko lori A: $T_B = T_A + 4$.
• Fomula ya umbali, kasi na muda ni $D = S \times T$, ambayo inaweza kuandikwa kama $T = \frac{D}{S}$.

Kwa lori A:

$$T_A = \frac{D}{S_A} = \frac{3120}{S_B + 5}$$

Kwa lori B:

$$T_B = \frac{D}{S_B} = \frac{3120}{S_B}$$

Badilisha milinganyo ya $T_A$ na $T_B$ kwenye $T_B = T_A + 4$:

$$\frac{3120}{S_B} = \frac{3120}{S_B + 5} + 4$$

Ili kuondoa sehemu, zidisha pande zote mbili za mlinganyo kwa $S_B(S_B + 5)$:

$$3120(S_B + 5) = 3120S_B + 4S_B(S_B + 5)$$

Panua mabano:

$$3120S_B + 15600 = 3120S_B + 4S_B^2 + 20S_B$$

Toa $3120S_B$ kutoka pande zote mbili:

$$15600 = 4S_B^2 + 20S_B$$

Panga upya mlinganyo katika fomu ya quadratic $aS_B^2 + bS_B + c = 0$:

$$4S_B^2 + 20S_B - 15600 = 0$$

Gawanya mlinganyo mzima kwa 4 ili kurahisisha:

$$S_B^2 + 5S_B - 3900 = 0$$

Tumia fomula ya quadratic $S_B = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, ambapo $a=1$, $b=5$, $c=-3900$:

$$S_B = \frac{-5 \pm \sqrt{5^2 - 4(1)(-3900)}}{2(1)}$$ $$S_B = \frac{-5 \pm \sqrt{25 + 15600}}{2}$$ $$S_B = \frac{-5 \pm \sqrt{15625}}{2}$$

Kokotoa mzizi wa mraba: $\sqrt{15625} = 125$.

$$S_B = \frac{-5 \pm 125}{2}$$

Kuna suluhisho mbili zinazowezekana:

$$S_{B1} = \frac{-5 + 125}{2} = \frac{120}{2} = 60$$ $$S_{B2} = \frac{-5 - 125}{2} = \frac{-130}{2} = -65$$

Kwa kuwa kasi haiwezi kuwa hasi, tunachagua thamani chanya.

Kasi ya lori B ni $\boxed{\text{60 km/h}}$.