A river, running between parallel banks, is 20 m wide - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 2

To use the trapezium rule:

• The width of each interval is $riangle x = 4$ m (since the x values are 0, 4, 8, 12, 16, and 20).
• The trapezium rule formula is given by: $A \approx \frac{\triangle x}{2} \left( y_0 + 2y_1 + 2y_2 + 2y_3 + 2y_4 + y_n \right)$

Substituting the calculated values: $A \approx \frac{4}{2} \left( 0.447 + 2(0.400) + 2(0.346) + 2(0.283) + 2(0.200) + 0 \right)$ $A \approx 2 \left( 0.447 + 0.800 + 0.692 + 0.566 + 0.400 + 0 \right)$ $A \approx 2 \cdot 3.905 = 7.810$

The estimated cross-sectional area is approximately $7.810 \, \text{m}^2$.

Given that the cross-sectional area is constant and the river flows uniformly at 2 m s⁻¹, the volume flow rate $Q$ is given by the formula: $Q = A \cdot v$ where:

• $A$ = cross-sectional area = $7.810 \, \text{m}^2$
• $v$ = velocity = $2 \, \text{m/s}$

So, $Q = 7.810 imes 2 = 15.620 \, \text{m}^3/s$

To convert this to volume per minute: $Q_{minute} = 15.620 imes 60 = sges930.800 \, \text{m}^3/min$

Rounding this to 3 significant figures, we get: $Q_{minute} \approx 930 \, \text{m}^3/min.$