Question $$ 6 \mathbf{c} $$ (WASSCE 2008 May/June) A cyclist starts a journey at $$ 10: 30 \mathrm{am} $$ and lans to get home 45 km away by 3.00 pm . At irst, he traveled for 2 hours at $$ 7 \frac{1}{2} \mathrm{kmh}^{-1} $$ and hen $$ x \mathrm{kmh}^{-1} $$ for the rest the time in order to rrive promptly at home. Find the value of $$ x $$.

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Let the total journey time be from $$10:30$$ am to $$3:00$$ pm. The total time is:

$$
3:00\text{ pm} - 10:30\text{ am} = 4.5 \text{ hours}
$$

The cyclist travels in two parts:
1. For the first 2 hours at a speed of $$7.5 \, \text{km/h}$$.
2. For the remaining $$4.5 - 2 = 2.5$$ hours at a speed of $$x \, \text{km/h}$$.

The distance covered during the first 2 hours is:

$$
\text{Distance}_1 = 7.5 \times 2 = 15 \, \text{km}
$$

The total distance is 45 km, so the distance to be covered in the remaining 2.5 hours is:

$$
\text{Distance}_2 = 45 - 15 = 30 \, \text{km}
$$

To find $$x$$, use the formula:

$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

Hence:

$$
30 = x \times 2.5
$$

Solving for $$x$$:

$$
x = \frac{30}{2.5} = 12 \, \text{km/h}
$$
The cyclist needs to travel at $$12 \, \text{km/h}$$ for the remaining 2.5 hours to arrive home on time.

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