Question

the table below gives te depth of water across a river measured at one metre intervals between banks. distance (m) 0 1 2 3 4 water depth (m) 0 0.5 1.6 0.9 0 use the trapezium rule to estimate the cross-sectional area of the river. a river hydrologist estimates that at the place where this cross sectional data was measured the average speed of water flow is 0.6m/s estimate the volum of water which passes this section of the river in one minute.

Answer 1

The table below gives the depth of water across a river measured at one metre intervals between banks. Distance (m) 0 1 2 3 4, water depth (m) 0 0.5 1.6 0.9 0. Use the trapezium rule to estimate the cross-sectional area of the river. A river hydrologist estimates that at the place where this cross sectional data was measured the average speed of water flow is $0.6\mathrm{m / s}$ . Estimate the volume of water which passes this section of the river in one minute.

Solution.

The cross-sectional area of the water is shown on the graph:

Total area is: $S = S_{1} + S_{2} + S_{3} + S_{4}$

Points of 1st triangle: $(0,0)$ , $(1,0)$ , $(1,0.5)$ .

Area of 1st triangle: $S_{1} = \frac{1}{2} \cdot (1 - 0) \cdot (0.5 - 0) = 0.25(m^{2})$

2nd trapezium points: $(1,0)$ , $(2,0)$ , $(1,0.5)$ , $(2,1.6)$ .

2nd trapezium area: $S_{2} = \frac{(0.5 - 0) + (1.6 - 0)}{2} \cdot (2 - 1) = 1.05(m^{2})$

3rd trapezium points: $(2,0)$ , $(3,0)$ , $(2,1.6)$ , $(3,0.9)$ .

3rd trapezium area: $S_{3} = \frac{(1.6 - 0) + (0.9 - 0)}{2} \cdot (3 - 2) = 1.25(m^{2})$

Points of 4th triangle: $(3,0)$ , $(4,0)$ , $(3,0.9)$ .

Area of 4th triangle: $S_{4} = \frac{1}{2} \cdot (4 - 3) \cdot (0.9 - 0) = 0.45(m^{2})$

So, total area is: $S = 0.25 + 1.05 + 1.25 + 0.45 = 3(m^2)$

The volume of water passes this section of the river in one minute is:

$$V = S \cdot v \cdot t = 3 m ^ {2} \cdot 0. 6 \frac {m}{s} \cdot 6 0 s = 1 0 8 (m ^ {3})$$

Answer: $S = 3m^2$ , $V = 108m^3$ .