A river, running between parallel banks, is 20 m wide - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 2

Using the trapezium rule, we estimate the area A under the curve:

$A = \frac{h}{2} \left( y_0 + 2y_1 + 2y_2 + 2y_3 + 2y_4 + y_5 \right)$

where h = 4 (the distance between x values), and the y values calculated above:

• $y_0 = 0.447$
• $y_1 = 0.400$
• $y_2 = 0.346$
• $y_3 = 0.283$
• $y_4 = 0.200$
• $y_5 = 0$

Substituting:

$A = \frac{4}{2} \left( 0.447 + 2(0.400) + 2(0.346) + 2(0.283) + 2(0.200) + 0 \right)$

Calculating this gives:

$A = 2 \left( 0.447 + 0.800 + 0.692 + 0.566 + 0.400 + 0 \right) = 2 \times 3.905 = 7.81 \text{ m}^2.$

The volume of water flowing per minute can be calculated using the formula:

$V = A \times ext{velocity}$

where

• $A = 7.81 \text{ m}^2$ (cross-sectional area), and
• velocity = 2 m/s.

Thus:

$V = 7.81 \text{ m}^2 \times 2 \text{ m/s} = 15.62 \text{ m}^3/ ext{s}$

To find the volume per minute, multiply by 60:

$V = 15.62 \text{ m}^3/ ext{s} \times 60 \text{ s} = 937.2 \text{ m}^3.$

Rounding to 3 significant figures, the final answer is:

$V \approx 937 \text{ m}^3.$