Calculate the total distance travelled by an object, given its velocity-time graph divided into four sections: a triangle, a rectangle, a trapezium, and another triangle.

• Section 1 (0 s to 6 s): A triangle.
• Section 2 (6 s to 9 s): A rectangle.
• Section 3 (9 s to 12 s): A trapezium.
• Section 4 (12 s to 18 s): A triangle.

• Section 1: Triangle (0 s to 6 s) Base ($b_1$) = $6 \text{ s} - 0 \text{ s} = 6 \text{ s}$ Height ($h_1$) = $20 \text{ m/s}$ (velocity at $t=6$ s) Area$_1 = \frac{1}{2} \times b_1 \times h_1 = \frac{1}{2} \times 6 \text{ s} \times 20 \text{ m/s} = 60 \text{ m}$

• Section 2: Rectangle (6 s to 9 s) Width ($w_2$) = $9 \text{ s} - 6 \text{ s} = 3 \text{ s}$ Height ($h_2$) = $20 \text{ m/s}$ (constant velocity) Area$_2 = w_2 \times h_2 = 3 \text{ s} \times 20 \text{ m/s} = 60 \text{ m}$

• Section 3: Trapezium (9 s to 12 s) Parallel side 1 ($v_a$) = $20 \text{ m/s}$ (velocity at $t=9$ s) Parallel side 2 ($v_b$) = $40 \text{ m/s}$ (velocity at $t=12$ s) Height ($h_3$) = $12 \text{ s} - 9 \text{ s} = 3 \text{ s}$ Area$_3 = \frac{1}{2} \times (v_a + v_b) \times h_3 = \frac{1}{2} \times (20 \text{ m/s} + 40 \text{ m/s}) \times 3 \text{ s}$ Area$_3 = \frac{1}{2} \times 60 \text{ m/s} \times 3 \text{ s} = 90 \text{ m}$

• Section 4: Triangle (12 s to 18 s) Base ($b_4$) = $18 \text{ s} - 12 \text{ s} = 6 \text{ s}$ Height ($h_4$) = $40 \text{ m/s}$ (velocity at $t=12$ s) Area$_4 = \frac{1}{2} \times b_4 \times h_4 = \frac{1}{2} \times 6 \text{ s} \times 40 \text{ m/s} = 120 \text{ m}$

Total Distance = Area$_1$ + Area$_2$ + Area$_3$ + Area$_4$
Total Distance = $60 \text{ m} + 60 \text{ m} + 90 \text{ m} + 120 \text{ m}$
Total Distance = $330 \text{ m}$