Calculate the acceleration of the particle in the first 10 seconds the value of t if it travelled 50 meters from 20th seconds the average speed for the whole journey

From $t=0$ s to $t=10$ s:
Initial velocity, $u = 30$ m/s
Final velocity, $v = 10$ m/s
Time interval, $\Delta t = 10$ s

Acceleration $a = \frac{v - u}{\Delta t}$

The acceleration of the particle in the first 10 seconds is $\boxed{\text{-2 m/s}^2}$.

2. The value of t if it travelled 50 meters from 20th seconds.

This segment is a triangle.
Initial velocity at $t=20$ s is 10 m/s.
Final velocity at $t=t$ s is 0 m/s.
The base of the triangle is $(t - 20)$ s.
The height of the triangle is 10 m/s.
The distance travelled is the area under this segment, which is given as 50 m.

Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$

Divide both sides by 5:

Add 20 to both sides:

The value of $t$ is $\boxed{\text{30 s}}$.

3. The average speed for the whole journey.

This is the total area under the graph from $t=0$ s to $t=30$ s.
The journey consists of three parts:

• Part 1 (0 to 10 s): A trapezium. Area$_1 = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$ $$\text{Area}_1 = \frac{1}{2} \times (30 \text{ m/s} + 10 \text{ m/s}) \times 10 \text{ s}$$ $$\text{Area}_1 = \frac{1}{2} \times 40 \text{ m/s} \times 10 \text{ s} = 200 \text{ m}$$
• Part 2 (10 to 20 s): A rectangle. Area$_2 = \text{length} \times \text{width}$ $$\text{Area}_2 = 10 \text{ m/s} \times (20 - 10) \text{ s} = 10 \text{ m/s} \times 10 \text{ s} = 100 \text{ m}$$
• Part 3 (20 to 30 s): A triangle. The distance for this part is given as 50 m in question 2. Area$_3 = 50$ m

Total distance = Area$_1$ + Area$_2$ + Area$_3$

Total time = $t = 30$ s (from question 2).

Average speed = $\frac{\text{Total distance travelled}}{\text{Total time taken}}$

The average speed for the whole journey is $\boxed{\text{11.67 m/s}}$.