📚 Year 10 AQA Physics: Interdisciplinary Integrated Question Practice | Year 10 AQA 物理：跨学科综合题型训练
In the AQA GCSE Physics specification, questions often cross topic boundaries and require you to apply knowledge from different areas of physics, as well as mathematical skills. This integrated question practice helps you build confidence in tackling problems that combine concepts such as energy, forces, electricity, and waves, while also using data analysis and experimental design skills. Working through these multi-step problems will deepen your understanding and prepare you thoroughly for your examinations.
在 AQA GCSE 物理课程中，试题往往跨越主题界线，要求你运用来自物理不同领域的知识以及数学技能。这种综合题型训练有助于你建立自信，解决结合了能量、力、电和波等概念的题目，同时需要使用数据分析与实验设计技能。通过练习这些多步问题进行训练，将加深你的理解，并为考试做好充分准备。
1. Rearranging Equations and Substituting Values | 方程重组与数值代入
tial because you will rarely be given numbers in exactly the order needed. For example, Ohm' ge to R = V / I. Practise moving terms until the desired quantity is isolated on one side.
许多物理问题需要你在代入数值之前重组方程。熟练掌握代数至关重要，因为提供的数字很少会恰好符合要求顺序。例如，欧姆定律给出的是 V = I R，但题目常常要求计算电阻，因此你必须将其重组为 R = V / I。请练习移项，直至所需物理量单独出现在等式的一边。
Always write the original equation, then show each algebraic step clearly. This method not only reduces mistakes but also earns method marks even if a final numerical error occurs. Once you have the rearranged form, substitute the values with their units and perform the calculation step by step.
始终写出原始方程，然后清晰地展示每一步代数运算。这种方法不仅能减少错误，即使最终答案出现数值错误，也能赢得方法分。得到重组形式后，代入包含单位的数值，逐步进行计算。
A typical question might combine charge and resistance:
一个典型题目可能会将电荷和电阻结合起来：
A lamp carries a current of 0.5 A when connected to a 12 V supply. Calculate its resistance and the charge that flows through it in 2 minutes.
一盏灯在接入12 V电源时通过电流0.5 A。计算它的电阻以及在2分钟内流过的电荷量。
Write Ohm's law: V = I R. Rearrange: R = V / I. Substitute: R = 12 / 0.5 = 24 Ω.
写出欧姆定律：V = I R。重组：R = V / I。代入：R = 12 / 0.5 = 24 Ω。
Use the charge equation Q = I t. Convert time: t = 2 × 60 = 120 s. Then Q = 0.5 × 120 = 60 C.
使用电荷方程 Q = I t。转换时间：t = 2 × 60 = 120 s。于是 Q = 0.5 × 120 = 60 C。
Check that units are consistent—time must be in seconds for charge calculations.
检查单位是否一致——计算电荷时时间必须使用秒。
2. Graph Interpretation and Gradient Calculations | 图表解释与斜率计算
Distance–time and velocity–time graphs appear frequently in AQA questions, often linked to real-world motion. The gradient of a distance–time graph gives speed; the gradient of a velocity–time graph gives acceleration. The area under a velocity–time graph represents displacement. You must be able to calculate gradients using Δy / Δx and find areas of simple shapes such as rectangles and triangles.
距离-时间图和速度-时间图常出现在 AQA 的考题中，并经常与真实运动联系。距离-时间图的斜率给出速率；速度-时间图的斜率给出加速度。速度-时间图下的面积表示位移。你必须能够使用 Δy / Δx 计算斜率，并求出矩形和三角形等简单形状的面积。
When interpreting a graph, first identify the type of motion in each section: constant speed, acceleration, or deceleration. Use the equations of motion alongside the graph to calculate missing quantities. A cross-topic link often appears when you combine graph information with energy calculations, for example using the final velocity to find kinetic energy.
解读图像时，首先识别每一段运动的类型：匀速、加速或减速。同时使用运动学方程和图像来计算缺失的物理量。当你在解题中将图像信息与能量计算相结合时（例如利用末速度求出动能），通常就会出现跨主题的联系。
Example: A cyclist travels at constant speed for 10 s covering 60 m, then accelerates uniformly for 20 s reaching a speed of 10 m/s. Sketch the velocity–time graph and calculate the total distance.
示例：一名自行车手匀速行驶10秒，前行60米，然后匀加速20秒，速度达到10 m/s。画出速度-时间图并计算总路程。
= 6 m/s for the first section. The graph shows a horizontal line from 0 to 10 s at 6 m/s.
第一阶段匀速 v = 60 / 10 = 6 m/s，图上0至10秒处为一条6 m/s的水平线。
Acceleration a = (10 – 6) / 20 = 0.2 m/s². The graph line slopes up from 6 m/s to 10 m/s between 10 s and 30 s.
加速度 a = (10 – 6) / 20 = 0.2 m/s²，在10至30秒之间，图线从6 m/s斜升至10 m/s。
ce = area under graph = (6 × 10) + area of trapezium = 60 + ½(6+10)×20 = 60 + 160 = 220 m.
路程 = 图下面积 = (6 × 10) + 梯形面积 = 60 + ½(6+10)×20 = 60 + 160 = 220 m。
3. Energy Transfers and Efficiency | 能量转移与效率
Efficiency questions blend multiple energy stores and often link electrical power with mechanical work. Efficiency is defined as (useful output energy / total input energy)
Published by TutorHao | Year 10 Physics Revision Series | aleveler.com
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