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P2 2023 Oct Q6

A Level / Edexcel / P2

IAL 2023 Oct Paper · Question 6

题目

Problem

A river is being studied.

At one particular place, the river is 15 m wide.

The depth, yy metres, of the river is measured at a point xx metres from one side of the river.

Figure 1 shows a plot of the cross-section of the river and the coordinate values (x,y)(x, y).

(a) Use the trapezium rule with all the yy values given in Figure 1 to estimate the cross-sectional area of the river.

(3)

The water in the river is modelled as flowing at a constant speed of 1.51.5 m s−1^{-1} across the whole of the cross-section.

(b) Use the model and the answer to part (a) to estimate the volume of water flowing through this section of the river each minute, giving your answer in m3^3 to 2 significant figures.

(2)

Assuming the model,

(c) state, giving a reason for your answer, whether your answer for part (b) is an overestimate or an underestimate of the true volume of water flowing through this section of the river each minute.

(1)
题目中文翻译

一条河流正在被研究。

在某一特定位置，河流宽15米。

河流的深度 yy（米）是在距河流一侧 xx（米）处测量的。

图1显示了河流横截面的图形以及坐标值 (x,y)(x, y)。

(a) 使用图1中给出的所有 yy 值，用梯形法则估算河流的横截面积。

（3分）

模型假设河水以恒定速度 1.51.5 m/s 流过整个横截面。

(b) 利用模型和(a)的答案，估算每分钟流过该河段的水的体积，答案以 m3^3 为单位，保留2位有效数字。

（2分）

假设模型成立，

(c) 指出(b)的答案是高估还是低估了真实值，并说明理由。

（1分）

解答

(a)

解法一

思路

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共五个等宽区间，宽度为 3 m；用全部给定深度代入梯形公式。

答题过程

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From Figure 1, the width of the river is 1515 m and the yy values are measured at equally spaced intervals. Therefore h=155=3h = \dfrac{15}{5} = 3.

Applying the trapezium rule:

A≈ 32[0+0+2(1.52+2.74 +3.12+3.08)]= 32×2×10.46= 31.38 m2\begin{align*} A\approx&\,\frac32\big[0+0+2(1.52+2.74\\ &\,\hspace{2pt}+3.12+3.08)\big]\\ =&\, \frac{3}{2} \times 2 \times 10.46 \\ =&\, 31.38 \text{ m}^2 \end{align*} A≈31.38 m2\boxed{A \approx 31.38 \text{ m}^2}

也可将两端三角形与中间三个梯形逐一相加：

A≈ 12×3×1.52 +32(1.52+2.74) +32(2.74+3.12) +32(3.12+3.08) +12×3×3.08= 31.38 m2.\begin{align*} A\approx&\,\frac12\times3\times1.52\\ &\,\hspace{2pt}+\frac32(1.52+2.74)\\ &\,\hspace{4pt}+\frac32(2.74+3.12)\\ &\,\hspace{6pt}+\frac32(3.12+3.08)\\ &\,\hspace{8pt}+\frac12\times3\times3.08\\ =&\,31.38\text{ m}^2. \end{align*}

(b)

解法一

思路

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模型中每分钟水流前进 90 m，用横截面积乘此长度得到体积，再按两位有效数字舍入。

答题过程

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The water flows at a constant speed of 1.51.5 m s−1^{-1}. The volume per minute is:

V= 31.38×1.5×60= 2824.2≈2800 m3(2 s.f.)\begin{align*} V =&\, 31.38 \times 1.5 \times 60 \\ =&\, 2824.2 \\ &\approx 2800 \text{ m}^3 \quad (2 \text{ s.f.}) \end{align*} V≈2800 m3\boxed{V \approx 2800 \text{ m}^3}

(c)

解法一

思路

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深度轴向下。梯形底边的直线近似比实际河床浅，所以梯形面积较小，体积也被低估。

答题过程

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This is an underestimate.

The depth axis points downwards. The straight bottom edges of the trapezia lie above the actual river bed, so each trapezium represents a shallower section and the total area of the trapezia is smaller than the true cross-sectional area. Under the constant-speed model, the volume estimate is therefore also too small.