A car's speed is graphed over 12 seconds. It travels 40m in the first 4 seconds. Find max speed and average speed for the whole journey.

The distance travelled is the area under the speed-time graph. For the first 4 seconds, the graph forms a triangle.
Let the maximum speed be $V_{max}$ (which is the height of the triangle).
The base of the triangle is 4 seconds.
The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

$$40 \text{ m} = \frac{1}{2} \times 4 \text{ s} \times V_{max}$$ $$40 = 2 V_{max}$$

$$V_{max} = \frac{40}{2}$$ $$V_{max} = \boxed{\text{20 m/s}}$$

(ii) Find the average speed for the whole journey.

The total distance is the total area under the speed-time graph from $t=0$ to $t=12$ seconds. This shape is a trapezium, which can be divided into three parts:

1. Triangle (0 to 4 s): Area = 40 m (given).
2. Rectangle (4 to 10 s): Base = $(10 - 4)$ s = 6 s, Height = $V_{max} = 20$ m/s. Area = $6 \text{ s} \times 20 \text{ m/s} = 120 \text{ m}$.
3. Triangle (10 to 12 s): Base = $(12 - 10)$ s = 2 s, Height = $V_{max} = 20$ m/s. Area = $\frac{1}{2} \times 2 \text{ s} \times 20 \text{ m/s} = 20 \text{ m}$. Total distance = $40 \text{ m} + 120 \text{ m} + 20 \text{ m} = 180 \text{ m}$.

Average speed is given by $\frac{\text{Total distance}}{\text{Total time}}$.
Total time = 12 seconds.

$$\text{Average speed} = \frac{180 \text{ m}}{12 \text{ s}}$$ $$\text{Average speed} = \boxed{\text{15 m/s}}$$

22. (b) Wafwila left Webuye at 8:00 am towards Mumias at an average speed of 90 km/hr. Nekesa also left Webuye at 8:21 am towards Mumias along the same road at an average speed of 97 km/hr.

(i) Determine the time Nekesa caught up with Wafwila.

Let $t$ be the time in hours that Wafwila has been travelling since 8:00 am.
Distance covered by Wafwila ($D_W$) = Speed $\times$ Time = $90t$.

Nekesa started 21 minutes later. Convert 21 minutes to hours:

$$21 \text{ minutes} = \frac{21}{60} \text{ hours} = 0.35 \text{ hours}$$

Nekesa's travel time = $(t - 0.35)$ hours.
Distance covered by Nekesa ($D_N$) = Speed $\times$ Time = $97(t - 0.35)$.

When Nekesa catches up with Wafwila, their distances from Webuye are equal:

$$D_W = D_N$$ $$90t = 97(t - 0.35)$$ $$90t = 97t - 97 \times 0.35$$ $$90t = 97t - 33.95$$ $$33.95 = 97t - 90t$$ $$33.95 = 7t$$ $$t = \frac{33.95}{7}$$ $$t = 4.85 \text{ hours}$$

$t = 4.85$ hours = 4 hours and $(0.85 \times 60)$ minutes.
$0.85 \times 60 = 51$ minutes.
So, Nekesa caught up 4 hours and 51 minutes after 8:00 am.
Time = 8:00 am + 4 hours 51 minutes = $\boxed{\text{12:51 pm}}$

(ii) Determine the distance from Webuye when Nekesa caught up with Wafwila.

Using Wafwila's distance equation:

$$D_W = 90t$$ $$D_W = 90 \times 4.85$$ $$D_W = \boxed{\text{436.5 km}}$$