89:"UserProgressProvider" 8a:I[651451,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$89"] 8b:I[196793,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"Gingerbread"] 8d:"TopicRecentlyViewed" 8e:I[716297,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$8d"] 8f:"RecordResourceResume" 90:I[647935,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$8f"] 91:"TopicQuestionBank" 92:I[426249,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$91"] 93:T4f7,To eliminate the unknown reaction force $R_C$, we take moments about the contact point $C$ (which is $5\text{ m}$ from $A$). - **Moment of weight $15g$:** The midpoint is $3\text{ m}$ from $A$, so it is $5 - 3 = 2\text{ m}$ from $C$. The perpendicular distance from $C$ to the line of action of the weight is $2\cos\alpha\text{ m}$. This force creates a clockwise moment: $$\text{Clockwise Moment} = 15g \times (2\cos\alpha)$$ - **Moment of normal reaction $R_A$:** The distance along the ladder from $A$ to $C$ is $5\text{ m}$. The perpendicular distance from $C$ to the vertical line of action of $R_A$ is $5\cos\alpha\text{ m}$. This force creates an anticlockwise moment: $$\text{Anticlockwise Moment}_1 = R_A \times (5\cos\alpha)$$ - **Moment of friction force $F_A$:** The perpendicular distance from $C$ to the horizontal line of action of $F_A$ is $5\sin\alpha\text{ m}$. This force creates a clockwise moment: $$\text{Clockwise Moment}_2 = F_A \times (5\sin\alpha)$$ Equating clockwise and anticlockwise moments about $C$: $$R_A (5\cos\alpha) = 15g (2\cos\alpha) + F_A (5\sin\alpha)$$ Substitute $\cos\alpha = 0.8$ and $\sin\alpha = 0.6$: $$R_A (5 \times 0.8) = 15g (2 \times 0.8) + F_A (5 \times 0.6)$$ $$4 R_A = 24g + 3 F_A$$ This is our **Equation 1**.94:T503,To eliminate the unknown reaction force $R_C$, we take moments about the contact point $C$ (which is $5\text{ m}$ from $A$). - **Moment of weight $15g$: ** The midpoint is $3\text{ m}$ from $A$, so it is $5 - 3 = 2\text{ m}$ from $C$. The perpendicular distance from $C$ to the line of action of the weight is $2\cos\alpha\text{ m}$. This force creates an anticlockwise moment: $$\text{Anticlockwise Moment} = 15g \times (2\cos\alpha)$$ - **Moment of normal reaction $R_A$: ** The distance along the ladder from $A$ to $C$ is $5\text{ m}$. The perpendicular distance from $C$ to the vertical line of action of $R_A$ is $5\cos\alpha\text{ m}$. This force creates a clockwise moment: $$\text{Clockwise Moment}_1 = R_A \times (5\cos\alpha)$$ - **Moment of friction force $F_A$: ** The perpendicular distance from $C$ to the horizontal line of action of $F_A$ is $5\sin\alpha\text{ m}$. This force creates an anticlockwise moment: $$\text{Anticlockwise Moment}_2 = F_A \times (5\sin\alpha)$$ Equating anticlockwise and clockwise moments about $C$: $$R_A (5\cos\alpha) = 15g (2\cos\alpha) + F_A (5\sin\alpha)$$ Substitute $\cos\alpha = 0.8$ and $\sin\alpha = 0.6$: $$R_A (5 \times 0.8) = 15g (2 \times 0.8) + F_A (5 \times 0.6)$$ $$4 R_A = 24g + 3 F_A$$ This is our **Equation 1**.72:["$","$L8a",null,{"bundle":{"topicIds":["1730ea45-e101-4896-b6e3-427de57efcfb","04531be1-e7bc-4dea-bb18-22c3291c085b","5b3192ef-5a82-4feb-991f-7eee65347afe","175d57a9-bbf4-4b09-8d52-92111a87d949","b45de59f-e0f1-45c2-b8bd-7570b0298f13"],"paperIds":[],"stageIds":[],"stageTopicIds":[],"stageAssessmentIds":[],"topicNoteIds":[]},"children":[["$","$L8b",null,{"feature":"question-bank"}],"$L8c",["$","$L8e",null,{"subjectCode":"a_level_maths__ocr","subjectName":"Maths","topicSlug":"mechanics","topicTitle":"3 Mechanics","sectionTitle":null,"activity":"worksheet","resourceTitle":"3 Mechanics Questions","resourceHref":"/a-level/maths/ocr/mechanics/questions/q1","saveTrail":false}],["$","$L90",null,{"href":"/a-level/maths/ocr/mechanics/questions/q1","label":"3 Mechanics","subjectCode":"a_level_maths__ocr","surface":"questions"}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"name\":\"Question 54 – 3 Mechanics - A Level Maths Questions\",\"description\":\"Practise 3 Mechanics A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"3 Mechanics\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/ocr/mechanics/questions/q54\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L92",null,{"backHref":"/a-level/maths/ocr/questions","basePath":"/a-level/maths/ocr/mechanics/questions","formulaUrl":null,"initialIndex":53,"initialQuestionWindow":[{"id":"209e721b-5d18-4f49-b42e-08860b3f57ee","revisionId":"dfc0e7fd-ed78-4d41-a73f-7b88cff2988b","questionNumber":null,"questionType":"LA","specification":"A window cleaner's ladder $AB$ has a mass of $15$ kg and a length of $6$ m. The ladder is modelled as a uniform rod. It rests with its lower end $A$ on a rough horizontal pavement and its upper part against a smooth horizontal rail at point $C$. Given $AC = 5$ m and the ladder is in limiting equilibrium. The ladder makes an angle $\\alpha$ with the horizontal, where $\\tan \\alpha = 3/4$. The coefficient of friction at the ground is $\\mu$.","difficultyBand":"medium","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"fb7bdb2d-8f11-4223-a688-96eae1a234d8","label":"a","content":"Calculate the value of $\\mu$.","metadata":null,"markscheme":"- Geometry: $\\cos \\alpha = 4/5, \\sin \\alpha = 3/5$ B1\n- Moments about $A$: $3(15g \\cos \\alpha) = 5 R_C$ M1\n- $R_C = \\frac{45g(0.8)}{5} = 7.2g \\approx 70.6$ N A1\n- Resolving vertically: $R_A + R_C \\cos \\alpha = 15g$ M1\n- $R_A = 15g - 7.2g(0.8) = 9.24g \\approx 90.6$ N A1\n- Resolving horizontally: $\\mu R_A = R_C \\sin \\alpha$ M1\n- $\\mu = \\frac{7.2g(0.6)}{9.24g} \\approx 0.47$ (2 sf) A2","marks":8,"order":0,"markingRubric":{"totalMarks":8,"points":[{"id":"p1","marks":1,"description":"State or use correct trigonometric ratios for $\\sin \\alpha$ and $\\cos \\alpha$.","code":"B"},{"id":"p2","marks":1,"description":"Formulate a moments equation about $A$ using weight and $R_C$.","code":"M"},{"id":"p3","marks":1,"description":"Calculate the normal reaction force at $C$.","code":"A"},{"id":"p4","marks":1,"description":"Formulate a vertical resolution equation.","code":"M"},{"id":"p5","marks":1,"description":"Calculate the normal reaction force at $A$.","code":"A"},{"id":"p6","marks":1,"description":"Formulate a horizontal resolution equation (friction = horizontal reaction).","code":"M"},{"id":"p7","marks":2,"description":"Obtain final value of $\\mu$ (accept 0.468).","code":"A"}],"finalAnswers":["0.47"],"generatedAt":"2026-05-06T14:04:42.731Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"moments_about_a","label":"Moments about A","steps":[{"type":"text","order":0,"title":"Analyze the trigonometry","content":"Let's first determine the values of $\\cos \\alpha$ and $\\sin \\alpha$ from the given information:\n\n$$\\tan \\alpha = \\frac{3}{4}$$\n\nUsing a right-angled triangle with an opposite side of $3$ and an adjacent side of $4$, the hypotenuse is:\n\n$$\\sqrt{3^2 + 4^2} = 5$$\n\nThus, we can find the exact trigonometric values:\n\n$$\\cos \\alpha = \\frac{4}{5} = 0.8$$\n\n$$\\sin \\alpha = \\frac{3}{5} = 0.6$$"},{"type":"text","order":1,"title":"Identify forces acting on ladder","content":"Let's list all the forces acting on the ladder $AB$:\n\n1. **Weight of the ladder ($W$):** Acts vertically downwards at the midpoint of the ladder (since it is a uniform rod of length $6\\text{ m}$), which is at a distance of $3\\text{ m}$ from $A$.\n$$W = 15g\\text{ N}$$\n\n2. **Reaction force at the smooth rail $C$ ($R_C$):** Since the rail is smooth, the reaction force $R_C$ acts perpendicular to the ladder at point $C$, which is at a distance of $5\\text{ m}$ from $A$.\n\n3. **Normal reaction force at the ground $A$ ($R_A$):** Acts vertically upwards at $A$.\n\n4. **Frictional force at the ground $A$ ($F_A$):** Since the ladder tends to slip outwards (to the left), the friction force $F_A$ acts horizontally to the right. Since the ladder is in limiting equilibrium, we have:\n$$F_A = \\mu R_A$$","highlights":[{"text":"mass of 15 kg","location":"specification"},{"text":"length of 6 m","location":"specification"},{"text":"AC = 5 m","location":"specification"}]},{"type":"text","order":2,"title":"Take moments about A","content":"Taking moments about the base of the ladder $A$ eliminates the forces $R_A$ and $F_A$ because their lines of action pass directly through $A$.\n\nFor rotational equilibrium, the clockwise moments must equal the anticlockwise moments:\n\n- **Clockwise moment** is due to the weight $15g$:\n$$\\text{Moment} = 15g \\times 3\\cos\\alpha$$\n\n- **Anticlockwise moment** is due to the reaction force $R_C$ acting perpendicular to the ladder at a distance of $5\\text{ m}$:\n$$\\text{Moment} = R_C \\times 5$$\n\nEquating these moments:\n$$5 R_C = 3(15g \\cos\\alpha)$$"},{"type":"text","order":3,"title":"Calculate reaction force at C","content":"Now we substitute $\\cos\\alpha = 0.8$ into our moment equation to solve for $R_C$:\n\n$$\\begin{aligned} 5 R_C &= 45g(0.8) \\\\ 5 R_C &= 36g \\\\ R_C &= 7.2g \\end{aligned}$$\n\nUsing $g = 9.8\\text{ m s}^{-2}$, we find:\n$$R_C = 7.2 \\times 9.8 = 70.56\\text{ N} \\approx 70.6\\text{ N}$$"},{"type":"text","order":4,"title":"Resolve forces vertically","content":"Next, we resolve all forces acting on the ladder vertically.\n\nThe vertical forces are:\n- The normal reaction $R_A$ acting vertically upwards.\n- The vertical component of $R_C$ acting upwards. Since $R_C$ is perpendicular to the ladder, it makes an angle of $\\alpha$ with the vertical, meaning its vertical component is $R_C \\cos\\alpha$.\n- The weight $15g$ acting vertically downwards.\n\nSetting up the equation for vertical equilibrium:\n$$R_A + R_C \\cos\\alpha = 15g$$\n\nSubstitute the known values ($R_C = 7.2g$ and $\\cos\\alpha = 0.8$):\n$$\\begin{aligned} R_A + 7.2g(0.8) &= 15g \\\\ R_A + 5.76g &= 15g \\\\ R_A &= 9.24g \\end{aligned}$$\n\nUsing $g = 9.8\\text{ m s}^{-2}$:\n$$R_A = 9.24 \\times 9.8 = 90.552\\text{ N} \\approx 90.6\\text{ N}$$"},{"type":"text","order":5,"title":"Resolve forces horizontally","content":"Now we resolve all forces horizontally.\n\nThe horizontal forces are:\n- The frictional force $F_A = \\mu R_A$ acting horizontally to the right.\n- The horizontal component of $R_C$ acting horizontally to the left, which is given by $R_C \\sin\\alpha$.\n\nSetting up the equation for horizontal equilibrium:\n$$\\mu R_A = R_C \\sin\\alpha$$"},{"type":"text","order":6,"title":"Calculate the coefficient of friction","content":"Substitute $R_A = 9.24g$, $R_C = 7.2g$, and $\\sin\\alpha = 0.6$ into the friction equation:\n\n$$\\mu (9.24g) = (7.2g)(0.6)$$\n\nNotice that $g$ cancels out from both sides:\n\n$$\\begin{aligned} \\mu (9.24) &= 4.32 \\\\ \\mu &= \\frac{4.32}{9.24} \\\\ \\mu &\\approx 0.46753 \\end{aligned}$$\n\nRounding to $2$ significant figures (which is standard when using $g = 9.8\\text{ m s}^{-2}$ in mechanics):\n\n$$\\mu \\approx 0.47$$"}]},{"id":"moments_about_c","label":"Moments about C","steps":[{"type":"text","order":0,"title":"Analyze the forces and geometry","content":"First, let's define the forces acting on the ladder and determine the trigonometric ratios of the angle $\\alpha$.\n\nGiven $\\tan \\alpha = \\frac{3}{4}$, we can use a right-angled triangle to find $\\sin \\alpha$ and $\\cos \\alpha$:\n$$\\cos \\alpha = \\frac{4}{5} = 0.8$$\n\n$$\\sin \\alpha = \\frac{3}{5} = 0.6$$\n\nThe forces acting on the uniform ladder $AB$ (of length $6\\text{ m}$ and mass $15\\text{ kg}$) are:\n1. The weight of the ladder, $W = 15g$, acting vertically downwards at the midpoint ($3\\text{ m}$ from $A$).\n2. The normal reaction at the ground $A$, $R_A$, acting vertically upwards.\n3. The frictional force at the ground $A$, $F_A = \\mu R_A$, acting horizontally towards the wall (preventing the ladder from slipping outward).\n4. The reaction force at the smooth rail $C$, $R_C$, acting perpendicular to the ladder at point $C$ ($5\\text{ m}$ from $A$).","highlights":[{"text":"mass of 15 kg","location":"specification"},{"text":"length of 6 m","location":"specification"},{"text":"AC = 5 m","location":"specification"}]},{"type":"text","order":1,"title":"Take moments about C","content":"$93"},{"type":"text","order":2,"title":"Resolve forces vertically","content":"Next, let's resolve forces vertically for the system in equilibrium.\n\nThe reaction force $R_C$ is perpendicular to the ladder, making an angle of $\\alpha$ with the vertical.\nTherefore, its vertical component is $R_C \\cos\\alpha$ pointing upwards.\n\nResolving vertically:\n$$R_A + R_C \\cos\\alpha = 15g$$\n\nThis is our **Equation 2**."},{"type":"text","order":3,"title":"Resolve forces horizontally","content":"Now resolve the forces horizontally.\n\nThe horizontal component of $R_C$ is $R_C \\sin\\alpha$ pointing to the left.\nThe frictional force $F_A = \\mu R_A$ points to the right.\n\nResolving horizontally:\n$$F_A = R_C \\sin\\alpha$$\n\nThis is our **Equation 3**."},{"type":"text","order":4,"title":"Relate reaction forces","content":"From **Equation 2** and **Equation 3**, we can express $R_C$ and substitute to find a relationship between $R_A$ and $F_A$.\n\nFrom **Equation 3**:\n$$R_C = \\frac{F_A}{\\sin\\alpha} = \\frac{F_A}{0.6}$$\n\nSubstitute this into **Equation 2**:\n$$R_A + \\left(\\frac{F_A}{0.6}\\right) \\cos\\alpha = 15g$$\n\n$$R_A + \\left(\\frac{F_A}{0.6}\\right) 0.8 = 15g$$\n\n$$R_A + \\frac{4}{3}F_A = 15g$$\n\nThis is our **Equation 4**."},{"type":"text","order":5,"title":"Solve for RA and FA","content":"We now have a system of two linear equations with $R_A$ and $F_A$:\n\n- **Equation 1**:\n$$4 R_A - 3 F_A = 24g$$\n\n- **Equation 4**:\n$$R_A + \\frac{4}{3}F_A = 15g$$\n\nMultiply **Equation 4** by 4:\n$$4 R_A + \\frac{16}{3}F_A = 60g$$\n\nSubtract **Equation 1** from this new equation to eliminate $R_A$:\n$$\\left(4 R_A + \\frac{16}{3}F_A\\right) - (4 R_A - 3 F_A) = 60g - 24g$$\n\n$$\\frac{25}{3}F_A = 36g$$\n\n$$F_A = \\frac{108g}{25} = 4.32g$$\n\nNow substitute $F_A = 4.32g$ back into **Equation 4** to find $R_A$:\n$$R_A + \\frac{4}{3}(4.32g) = 15g$$\n\n$$R_A + 5.76g = 15g$$\n\n$$R_A = 9.24g$$"},{"type":"text","order":6,"title":"Calculate coefficient of friction","content":"Since the ladder is in limiting equilibrium, the frictional force is at its maximum value:\n$$F_A = \\mu R_A$$\n\nSubstitute our values for $F_A$ and $R_A$:\n$$\\mu = \\frac{F_A}{R_A} = \\frac{4.32g}{9.24g}$$\n\nSimplifying this fraction:\n$$\\mu = \\frac{4.32}{9.24} = \\frac{108}{231} = \\frac{36}{77} \\approx 0.47$$\n\nRounding to $2$ significant figures gives:\n$$\\mu \\approx 0.47$$"}]}],"generatedAt":"2026-06-09T02:01:32.349Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"fb7bdb2d-8f11-4223-a688-96eae1a234d8:moments_about_a","label":"Moments about A","steps":[{"type":"text","order":0,"title":"Analyze the trigonometry","content":"Let's first determine the values of $\\cos \\alpha$ and $\\sin \\alpha$ from the given information:\n\n$$\\tan \\alpha = \\frac{3}{4}$$\n\nUsing a right-angled triangle with an opposite side of $3$ and an adjacent side of $4$, the hypotenuse is:\n\n$$\\sqrt{3^2 + 4^2} = 5$$\n\nThus, we can find the exact trigonometric values:\n\n$$\\cos \\alpha = \\frac{4}{5} = 0.8$$\n\n$$\\sin \\alpha = \\frac{3}{5} = 0.6$$"},{"type":"text","order":1,"title":"Identify forces acting on ladder","content":"Let's list all the forces acting on the ladder $AB$:\n\n1. **Weight of the ladder ($W$):** Acts vertically downwards at the midpoint of the ladder (since it is a uniform rod of length $6\\text{ m}$), which is at a distance of $3\\text{ m}$ from $A$.\n$$W = 15g\\text{ N}$$\n\n2. **Reaction force at the smooth rail $C$ ($R_C$):** Since the rail is smooth, the reaction force $R_C$ acts perpendicular to the ladder at point $C$, which is at a distance of $5\\text{ m}$ from $A$.\n\n3. **Normal reaction force at the ground $A$ ($R_A$):** Acts vertically upwards at $A$.\n\n4. **Frictional force at the ground $A$ ($F_A$):** Since the ladder tends to slip outwards (to the left), the friction force $F_A$ acts horizontally to the right. Since the ladder is in limiting equilibrium, we have:\n$$F_A = \\mu R_A$$","highlights":[{"text":"mass of 15 kg","location":"specification"},{"text":"length of 6 m","location":"specification"},{"text":"AC = 5 m","location":"specification"}]},{"type":"text","order":2,"title":"Take moments about A","content":"Taking moments about the base of the ladder $A$ eliminates the forces $R_A$ and $F_A$ because their lines of action pass directly through $A$.\n\nFor rotational equilibrium, the clockwise moments must equal the anticlockwise moments:\n\n- **Clockwise moment** is due to the weight $15g$:\n$$\\text{Moment} = 15g \\times 3\\cos\\alpha$$\n\n- **Anticlockwise moment** is due to the reaction force $R_C$ acting perpendicular to the ladder at a distance of $5\\text{ m}$:\n$$\\text{Moment} = R_C \\times 5$$\n\nEquating these moments:\n$$5 R_C = 3(15g \\cos\\alpha)$$"},{"type":"text","order":3,"title":"Calculate reaction force at C","content":"Now we substitute $\\cos\\alpha = 0.8$ into our moment equation to solve for $R_C$:\n\n$$\\begin{aligned} 5 R_C &= 45g(0.8) \\\\ 5 R_C &= 36g \\\\ R_C &= 7.2g \\end{aligned}$$\n\nUsing $g = 9.8\\text{ m s}^{-2}$, we find:\n$$R_C = 7.2 \\times 9.8 = 70.56\\text{ N} \\approx 70.6\\text{ N}$$"},{"type":"text","order":4,"title":"Resolve forces vertically","content":"Next, we resolve all forces acting on the ladder vertically.\n\nThe vertical forces are:\n- The normal reaction $R_A$ acting vertically upwards.\n- The vertical component of $R_C$ acting upwards. Since $R_C$ is perpendicular to the ladder, it makes an angle of $\\alpha$ with the vertical, meaning its vertical component is $R_C \\cos\\alpha$.\n- The weight $15g$ acting vertically downwards.\n\nSetting up the equation for vertical equilibrium:\n$$R_A + R_C \\cos\\alpha = 15g$$\n\nSubstitute the known values ($R_C = 7.2g$ and $\\cos\\alpha = 0.8$):\n$$\\begin{aligned} R_A + 7.2g(0.8) &= 15g \\\\ R_A + 5.76g &= 15g \\\\ R_A &= 9.24g \\end{aligned}$$\n\nUsing $g = 9.8\\text{ m s}^{-2}$:\n$$R_A = 9.24 \\times 9.8 = 90.552\\text{ N} \\approx 90.6\\text{ N}$$"},{"type":"text","order":5,"title":"Resolve forces horizontally","content":"Now we resolve all forces horizontally.\n\nThe horizontal forces are:\n- The frictional force $F_A = \\mu R_A$ acting horizontally to the right.\n- The horizontal component of $R_C$ acting horizontally to the left, which is given by $R_C \\sin\\alpha$.\n\nSetting up the equation for horizontal equilibrium:\n$$\\mu R_A = R_C \\sin\\alpha$$"},{"type":"text","order":6,"title":"Calculate the coefficient of friction","content":"Substitute $R_A = 9.24g$, $R_C = 7.2g$, and $\\sin\\alpha = 0.6$ into the friction equation:\n\n$$\\mu (9.24g) = (7.2g)(0.6)$$\n\nNotice that $g$ cancels out from both sides:\n\n$$\\begin{aligned} \\mu (9.24) &= 4.32 \\\\ \\mu &= \\frac{4.32}{9.24} \\\\ \\mu &\\approx 0.46753 \\end{aligned}$$\n\nRounding to $2$ significant figures (which is standard when using $g = 9.8\\text{ m s}^{-2}$ in mechanics):\n\n$$\\mu \\approx 0.47$$"}]},{"id":"fb7bdb2d-8f11-4223-a688-96eae1a234d8:moments_about_c","label":"Moments about C","steps":[{"type":"text","order":0,"title":"Analyze the forces and geometry","content":"First, let's define the forces acting on the ladder and determine the trigonometric ratios of the angle $\\alpha$.\n\nGiven $\\tan \\alpha = \\frac{3}{4}$, we can use a right-angled triangle to find $\\sin \\alpha$ and $\\cos \\alpha$:\n$$\\cos \\alpha = \\frac{4}{5} = 0.8$$\n\n$$\\sin \\alpha = \\frac{3}{5} = 0.6$$\n\nThe forces acting on the uniform ladder $AB$ (of length $6\\text{ m}$ and mass $15\\text{ kg}$) are:\n1. The weight of the ladder, $W = 15g$, acting vertically downwards at the midpoint ($3\\text{ m}$ from $A$).\n2. The normal reaction at the ground $A$, $R_A$, acting vertically upwards.\n3. The frictional force at the ground $A$, $F_A = \\mu R_A$, acting horizontally towards the wall (preventing the ladder from slipping outward).\n4. The reaction force at the smooth rail $C$, $R_C$, acting perpendicular to the ladder at point $C$ ($5\\text{ m}$ from $A$).","highlights":[{"text":"mass of 15 kg","location":"specification"},{"text":"length of 6 m","location":"specification"},{"text":"AC = 5 m","location":"specification"}]},{"type":"text","order":1,"title":"Take moments about C","content":"$94"},{"type":"text","order":2,"title":"Resolve forces vertically","content":"Next, let's resolve forces vertically for the system in equilibrium.\n\nThe reaction force $R_C$ is perpendicular to the ladder, making an angle of $\\alpha$ with the vertical.\nTherefore, its vertical component is $R_C \\cos\\alpha$ pointing upwards.\n\nResolving vertically:\n$$R_A + R_C \\cos\\alpha = 15g$$\n\nThis is our **Equation 2**."},{"type":"text","order":3,"title":"Resolve forces horizontally","content":"Now resolve the forces horizontally.\n\nThe horizontal component of $R_C$ is $R_C \\sin\\alpha$ pointing to the left.\nThe frictional force $F_A = \\mu R_A$ points to the right.\n\nResolving horizontally:\n$$F_A = R_C \\sin\\alpha$$\n\nThis is our **Equation 3**."},{"type":"text","order":4,"title":"Relate reaction forces","content":"From **Equation 2** and **Equation 3**, we can express $R_C$ and substitute to find a relationship between $R_A$ and $F_A$.\n\nFrom **Equation 3**:\n$$R_C = \\frac{F_A}{\\sin\\alpha} = \\frac{F_A}{0.6}$$\n\nSubstitute this into **Equation 2**:\n$$R_A + \\left(\\frac{F_A}{0.6}\\right) \\cos\\alpha = 15g$$\n\n$$R_A + \\left(\\frac{F_A}{0.6}\\right) 0.8 = 15g$$\n\n$$R_A + \\frac{4}{3}F_A = 15g$$\n\nThis is our **Equation 4**."},{"type":"text","order":5,"title":"Solve for RA and FA","content":"We now have a system of two linear equations with $R_A$ and $F_A$:\n\n- **Equation 1**:\n$$4 R_A - 3 F_A = 24g$$\n\n- **Equation 4**:\n$$R_A + \\frac{4}{3}F_A = 15g$$\n\nMultiply **Equation 4** by 4:\n$$4 R_A + \\frac{16}{3}F_A = 60g$$\n\nSubtract **Equation 1** from this new equation to eliminate $R_A$:\n$$\\left(4 R_A + \\frac{16}{3}F_A\\right) - (4 R_A - 3 F_A) = 60g - 24g$$\n\n$$\\frac{25}{3}F_A = 36g$$\n\n$$F_A = \\frac{108g}{25} = 4.32g$$\n\nNow substitute $F_A = 4.32g$ back into **Equation 4** to find $R_A$:\n$$R_A + \\frac{4}{3}(4.32g) = 15g$$\n\n$$R_A + 5.76g = 15g$$\n\n$$R_A = 9.24g$$"},{"type":"text","order":6,"title":"Calculate coefficient of friction","content":"Since the ladder is in limiting equilibrium, the frictional force is at its maximum value:\n$$F_A = \\mu R_A$$\n\nSubstitute our values for $F_A$ and $R_A$:\n$$\\mu = \\frac{F_A}{R_A} = \\frac{4.32g}{9.24g}$$\n\nSimplifying this fraction:\n$$\\mu = \\frac{4.32}{9.24} = \\frac{108}{231} = \\frac{36}{77} \\approx 0.47$$\n\nRounding to $2$ significant figures gives:\n$$\\mu \\approx 0.47$$"}]}],"generatedAt":"2026-06-09T02:01:53.493Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"22588e43-3dd2-493b-a0ea-92169432fe33","revisionId":"ec06142d-5cc9-418f-ba2f-4279605e2755","questionNumber":null,"questionType":"LA","specification":"A hockey puck $Q$ slides across smooth ice with constant acceleration. At time $t = 0$, $Q$ is moving in the direction $(-8\\mathbf{i} + 6\\mathbf{j})$ with speed $20 \\text{ ms}^{-1}$. After 8 seconds, its velocity is $(4\\mathbf{i} + 18\\mathbf{j}) \\text{ ms}^{-1}$.","difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"b685de5f-a2be-4a55-9d09-2b851e81c203","label":"a","content":"Find the acceleration of $Q$.","metadata":null,"markscheme":"- Find the magnitude of the direction vector: $\\sqrt{(-8)^2 + 6^2} = 10$ M1\n- Determine initial velocity $\\mathbf{u}$: $\\mathbf{u} = \\frac{20}{10}(-8\\mathbf{i} + 6\\mathbf{j}) = -16\\mathbf{i} + 12\\mathbf{j}$ A1\n- Use $\\mathbf{v} = \\mathbf{u} + \\mathbf{a}t$: $4\\mathbf{i} + 18\\mathbf{j} = (-16\\mathbf{i} + 12\\mathbf{j}) + 8\\mathbf{a}$ M1\n- Rearrange for $8\\mathbf{a}$: $8\\mathbf{a} = 20\\mathbf{i} + 6\\mathbf{j}$ M1\n- Final answer: $\\mathbf{a} = 2.5\\mathbf{i} + 0.75\\mathbf{j}$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Calculate magnitude of initial direction vector (-8, 6).","code":"M"},{"id":"p2","marks":1,"description":"Correct initial velocity vector u = -16i + 12j.","code":"A"},{"id":"p3","marks":1,"description":"Substitute vectors into v = u + at.","code":"M"},{"id":"p4","marks":1,"description":"Rearrange equation to find the vector for 8a.","code":"M"},{"id":"p5","marks":1,"description":"Divide by t to find final acceleration vector.","code":"A"}],"finalAnswers":["2.5\\mathbf{i} + 0.75\\mathbf{j}"],"generatedAt":"2026-05-06T14:03:42.848Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"vector_kinematics","label":"Vector Kinematics","steps":[{"type":"text","order":0,"title":"Find the unit direction vector","content":"To find the initial velocity vector $\\mathbf{u}$, we first need to scale the direction vector $(-8\\mathbf{i} + 6\\mathbf{j})$ so that its magnitude matches the given speed of $20\\text{ ms}^{-1}$.\n\nWe start by calculating the magnitude of this direction vector:\n\n$$\\text{Magnitude} = \\sqrt{(-8)^2 + 6^2} = \\sqrt{64 + 36} = \\sqrt{100} = 10$$"},{"type":"text","order":1,"title":"Calculate initial velocity vector","content":"Now, we scale the direction vector to find the initial velocity vector $\\mathbf{u}$. We divide the direction vector by its magnitude to obtain a unit vector, and then multiply by the speed, which is $20\\text{ ms}^{-1}$:\n\n$$\\mathbf{u} = \\frac{20}{10}(-8\\mathbf{i} + 6\\mathbf{j})$$\n\n$$\\mathbf{u} = 2(-8\\mathbf{i} + 6\\mathbf{j}) = -16\\mathbf{i} + 12\\mathbf{j}$$"},{"type":"text","order":2,"title":"Set up kinematic equation","content":"Since the hockey puck moves with constant acceleration, we can use the 2D vector kinematic equation:\n\n$$\\mathbf{v} = \\mathbf{u} + \\mathbf{a}t$$\n\nWe are given:\n- Final velocity $\\mathbf{v} = 4\\mathbf{i} + 18\\mathbf{j}$\n- Time $t = 8\\text{ s}$\n\nSubstituting our known values into the formula:\n\n$$4\\mathbf{i} + 18\\mathbf{j} = (-16\\mathbf{i} + 12\\mathbf{j}) + 8\\mathbf{a}$$"},{"type":"text","order":3,"title":"Solve for acceleration vector","content":"We can now rearrange the equation to find $8\\mathbf{a}$:\n\n$$8\\mathbf{a} = (4\\mathbf{i} + 18\\mathbf{j}) - (-16\\mathbf{i} + 12\\mathbf{j})$$\n\n$$8\\mathbf{a} = 20\\mathbf{i} + 6\\mathbf{j}$$\n\nDivide by 8 to get the acceleration vector $\\mathbf{a}$:\n\n$$\\mathbf{a} = 2.5\\mathbf{i} + 0.75\\mathbf{j}\\text{ ms}^{-2}$$"}]},{"id":"component_analysis","label":"Component-wise Analysis","steps":[{"type":"text","order":0,"title":"Analyze the direction vector","content":"The question tells us that at $t = 0$, the puck is moving in the direction of the vector $-8\\mathbf{i} + 6\\mathbf{j}$.\n\nTo find the actual initial velocity component-wise, we first need to determine the magnitude of this direction vector. Using the Pythagorean theorem:\n\n$$\\text{Magnitude} = \\sqrt{(-8)^2 + 6^2} = \\sqrt{64 + 36} = 10$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\mathbf{u} = \\frac{\\htmlData{anchor=speed}{\\color{@sky}{20}}}{\\htmlData{anchor=mag}{\\color{@pink}{10}}}(\\htmlData{anchor=dir}{\\color{@purple}{-8\\mathbf{i} + 6\\mathbf{j}}})$"},{"latex":"$$\\mathbf{u} = \\htmlData{anchor=scale}{\\color{@teal}{2}}(-8\\mathbf{i} + 6\\mathbf{j}) = \\htmlData{anchor=u-vector}{\\color{@orange}{-16\\mathbf{i} + 12\\mathbf{j}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"speed"},{"color":"pink","style":"text-color","anchor":"mag"},{"color":"teal","style":"text-color","anchor":"scale"},{"color":"orange","style":"box","anchor":"u-vector"}],"connections":[{"to":"scale","from":"speed","color":"sky","label":"speed ratio","style":"solid"}]},"order":1,"title":"Find the initial velocity components","description":"We scale the direction vector to match the actual initial speed of $20\\text{ ms}^{-1}$. Let's calculate the horizontal ($u_x$) and vertical ($u_y$) components of the initial velocity $\\mathbf{u}$."},{"type":"text","order":2,"title":"Resolve the final velocity","content":"We are given that after $t = 8$ seconds, the final velocity is $\\mathbf{v} = 4\\mathbf{i} + 18\\mathbf{j}$.\n\nWe can split this into horizontal and vertical final velocity components:\n- Horizontal final velocity: $v_x = 4\\text{ ms}^{-1}$\n- Vertical final velocity: $v_y = 18\\text{ ms}^{-1}$"},{"type":"text","order":3,"title":"Solve for horizontal acceleration","content":"Using the scalar constant acceleration formula $v = u + at$ for the horizontal component ($x$):\n\n$$v_x = u_x + a_x t$$\n\n$$4 = -16 + a_x(8)$$\n\n$$20 = 8a_x$$\n\n$$a_x = 2.5\\text{ ms}^{-2}$$"},{"type":"text","order":4,"title":"Solve for vertical acceleration","content":"Similarly, we apply the scalar formula $v = u + at$ to the vertical component ($y$):\n\n$$v_y = u_y + a_y t$$\n\n$$18 = 12 + a_y(8)$$\n\n$$6 = 8a_y$$\n\n$$a_y = 0.75\\text{ ms}^{-2}$$"},{"type":"text","order":5,"title":"Combine components into a vector","content":"Now we combine our horizontal and vertical acceleration components back into a single vector expression:\n\n$$\\mathbf{a} = a_x\\mathbf{i} + a_y\\mathbf{j}$$\n\n$$\\mathbf{a} = 2.5\\mathbf{i} + 0.75\\mathbf{j}$$"}]}],"generatedAt":"2026-06-09T03:48:40.045Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"b685de5f-a2be-4a55-9d09-2b851e81c203:vector_kinematics","label":"Vector Kinematics","steps":[{"type":"text","order":0,"title":"Find the unit direction vector","content":"To find the initial velocity vector $\\mathbf{u}$, we first need to scale the direction vector $(-8\\mathbf{i} + 6\\mathbf{j})$ so that its magnitude matches the given speed of $20\\text{ ms}^{-1}$.\n\nWe start by calculating the magnitude of this direction vector:\n\n$$\\text{Magnitude} = \\sqrt{(-8)^2 + 6^2} = \\sqrt{64 + 36} = \\sqrt{100} = 10$$"},{"type":"text","order":1,"title":"Calculate initial velocity vector","content":"Now, we scale the direction vector to find the initial velocity vector $\\mathbf{u}$. We divide the direction vector by its magnitude to obtain a unit vector, and then multiply by the speed, which is $20\\text{ ms}^{-1}$:\n\n$$\\mathbf{u} = \\frac{20}{10}(-8\\mathbf{i} + 6\\mathbf{j})$$\n\n$$\\mathbf{u} = 2(-8\\mathbf{i} + 6\\mathbf{j}) = -16\\mathbf{i} + 12\\mathbf{j}$$"},{"type":"text","order":2,"title":"Set up kinematic equation","content":"Since the hockey puck moves with constant acceleration, we can use the 2D vector kinematic equation:\n\n$$\\mathbf{v} = \\mathbf{u} + \\mathbf{a}t$$\n\nWe are given:\n- Final velocity $\\mathbf{v} = 4\\mathbf{i} + 18\\mathbf{j}$\n- Time $t = 8\\text{ s}$\n\nSubstituting our known values into the formula:\n\n$$4\\mathbf{i} + 18\\mathbf{j} = (-16\\mathbf{i} + 12\\mathbf{j}) + 8\\mathbf{a}$$"},{"type":"text","order":3,"title":"Solve for acceleration vector","content":"We can now rearrange the equation to find $8\\mathbf{a}$:\n\n$$8\\mathbf{a} = (4\\mathbf{i} + 18\\mathbf{j}) - (-16\\mathbf{i} + 12\\mathbf{j})$$\n\n$$8\\mathbf{a} = 20\\mathbf{i} + 6\\mathbf{j}$$\n\nDivide by 8 to get the acceleration vector $\\mathbf{a}$:\n\n$$\\mathbf{a} = 2.5\\mathbf{i} + 0.75\\mathbf{j}\\text{ ms}^{-2}$$"}]},{"id":"b685de5f-a2be-4a55-9d09-2b851e81c203:component_analysis","label":"Component-wise Analysis","steps":[{"type":"text","order":0,"title":"Analyze the direction vector","content":"The question tells us that at $t = 0$, the puck is moving in the direction of the vector $-8\\mathbf{i} + 6\\mathbf{j}$.\n\nTo find the actual initial velocity component-wise, we first need to determine the magnitude of this direction vector. Using the Pythagorean theorem:\n\n$$\\text{Magnitude} = \\sqrt{(-8)^2 + 6^2} = \\sqrt{64 + 36} = 10$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\mathbf{u} = \\frac{\\htmlData{anchor=speed}{\\color{@sky}{20}}}{\\htmlData{anchor=mag}{\\color{@pink}{10}}}(\\htmlData{anchor=dir}{\\color{@purple}{-8\\mathbf{i} + 6\\mathbf{j}}})$"},{"latex":"$$\\mathbf{u} = \\htmlData{anchor=scale}{\\color{@teal}{2}}(-8\\mathbf{i} + 6\\mathbf{j}) = \\htmlData{anchor=u-vector}{\\color{@orange}{-16\\mathbf{i} + 12\\mathbf{j}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"speed"},{"color":"pink","style":"text-color","anchor":"mag"},{"color":"teal","style":"text-color","anchor":"scale"},{"color":"orange","style":"box","anchor":"u-vector"}],"connections":[{"to":"scale","from":"speed","color":"sky","label":"speed ratio","style":"solid"}]},"order":1,"title":"Find the initial velocity components","description":"We scale the direction vector to match the actual initial speed of $20\\text{ ms}^{-1}$. Let's calculate the horizontal ($u_x$) and vertical ($u_y$) components of the initial velocity $\\mathbf{u}$."},{"type":"text","order":2,"title":"Resolve the final velocity","content":"We are given that after $t = 8$ seconds, the final velocity is $\\mathbf{v} = 4\\mathbf{i} + 18\\mathbf{j}$.\n\nWe can split this into horizontal and vertical final velocity components:\n- Horizontal final velocity: $v_x = 4\\text{ ms}^{-1}$\n- Vertical final velocity: $v_y = 18\\text{ ms}^{-1}$"},{"type":"text","order":3,"title":"Solve for horizontal acceleration","content":"Using the scalar constant acceleration formula $v = u + at$ for the horizontal component ($x$):\n\n$$v_x = u_x + a_x t$$\n\n$$4 = -16 + a_x(8)$$\n\n$$20 = 8a_x$$\n\n$$a_x = 2.5\\text{ ms}^{-2}$$"},{"type":"text","order":4,"title":"Solve for vertical acceleration","content":"Similarly, we apply the scalar formula $v = u + at$ to the vertical component ($y$):\n\n$$v_y = u_y + a_y t$$\n\n$$18 = 12 + a_y(8)$$\n\n$$6 = 8a_y$$\n\n$$a_y = 0.75\\text{ ms}^{-2}$$"},{"type":"text","order":5,"title":"Combine components into a vector","content":"Now we combine our horizontal and vertical acceleration components back into a single vector expression:\n\n$$\\mathbf{a} = a_x\\mathbf{i} + a_y\\mathbf{j}$$\n\n$$\\mathbf{a} = 2.5\\mathbf{i} + 0.75\\mathbf{j}$$"}]}],"generatedAt":"2026-06-09T03:49:06.621Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"2dfcf88b-cc51-4c05-9864-21e3131acd36","revisionId":"8ddcb16b-3059-4a59-b10d-5e13434c191f","questionNumber":null,"questionType":"LA","specification":"A toy boat $P$ moves across a pond with constant velocity. Initially $P$ is at the point $A$ with position vector $(-2\\mathbf{i} + 7\\mathbf{j}) \\text{ m}$. At time $t = 2$ seconds, $P$ is at the point $B$ with position vector $(8\\mathbf{i} - 17\\mathbf{j}) \\text{ m}$.","difficultyBand":"medium","tier":null,"totalMarks":7,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"492a1650-fea3-4b56-9c84-f2d49535af73","label":"a","content":"Find the velocity of $P$.","metadata":null,"markscheme":"- Use $\\mathbf{r} = \\mathbf{r}_0 + \\mathbf{v}t$: $8\\mathbf{i} - 17\\mathbf{j} = (-2\\mathbf{i} + 7\\mathbf{j}) + 2\\mathbf{v}$ M1\n- Subtract initial position: $10\\mathbf{i} - 24\\mathbf{j} = 2\\mathbf{v}$ M1\n- Final answer: $\\mathbf{v} = 5\\mathbf{i} - 12\\mathbf{j}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Substitute initial position, position at $t=2$, and $t=2$ into motion equation.","code":"M"},{"id":"p2","marks":1,"description":"Determine the total displacement $10\\mathbf{i} - 24\\mathbf{j}$.","code":"M"},{"id":"p3","marks":1,"description":"State correct velocity vector $5\\mathbf{i} - 12\\mathbf{j}$.","code":"A"}],"finalAnswers":["5\\mathbf{i} - 12\\mathbf{j}"],"generatedAt":"2026-05-06T14:03:50.806Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"vector_equation","label":"Vector Equation Approach","steps":[{"type":"text","order":0,"title":"Identify the vector equation","content":"For a particle moving with constant velocity, we can use the vector equation of a straight line:\n\n$$\\mathbf{r} = \\mathbf{r}_0 + \\mathbf{v}t$$\n\nwhere:\n- $\\mathbf{r}_0$ is the initial position vector.\n- $\\mathbf{v}$ is the constant velocity vector.\n- $\\mathbf{r}$ is the position vector at time $t$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\mathbf{r}_0 = \\htmlData{anchor=r0-val}{\\color{@sky}{-2\\mathbf{i} + 7\\mathbf{j}}}, \\; \\mathbf{r} = \\htmlData{anchor=r-val}{\\color{@amber}{8\\mathbf{i} - 17\\mathbf{j}}}, \\; t = \\htmlData{anchor=t-val}{\\color{@purple}{2}}$"},{"latex":"$$\\htmlData{anchor=r-eq}{\\color{@amber}{\\mathbf{r}}} = \\htmlData{anchor=r0-eq}{\\color{@sky}{\\mathbf{r}_0}} + \\mathbf{v}\\htmlData{anchor=t-eq}{\\color{@purple}{t}}$"},{"latex":"$$\\color{@amber}{8\\mathbf{i} - 17\\mathbf{j}} = (\\color{@sky}{-2\\mathbf{i} + 7\\mathbf{j}}) + 2\\mathbf{v}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"r0-val"},{"color":"amber","style":"text-color","anchor":"r-val"},{"color":"purple","style":"text-color","anchor":"t-val"}],"connections":[{"to":"r0-eq","from":"r0-val","color":"sky","label":"initial position","style":"solid"},{"to":"r-eq","from":"r-val","color":"amber","label":"final position","style":"solid"},{"to":"t-eq","from":"t-val","color":"purple","label":"time elapsed","style":"solid"}]},"order":1,"title":"Substitute the given values","description":"We substitute the given initial position vector, the final position vector, and the time into our equation."},{"type":"text","order":2,"title":"Isolate the velocity term","content":"To solve for $\\mathbf{v}$, we first subtract the initial position vector $(-2\\mathbf{i} + 7\\mathbf{j})$ from both sides to group the $\\mathbf{i}$ and $\\mathbf{j}$ components together:\n\n$$2\\mathbf{v} = (8\\mathbf{i} - 17\\mathbf{j}) - (-2\\mathbf{i} + 7\\mathbf{j})$$\n\n$$2\\mathbf{v} = (8 - (-2))\\mathbf{i} + (-17 - 7)\\mathbf{j}$$\n\n$$2\\mathbf{v} = 10\\mathbf{i} - 24\\mathbf{j}$$"},{"type":"text","order":3,"title":"Solve for the velocity vector","content":"Finally, we divide both sides of the equation by $2$ to isolate $\\mathbf{v}$:\n\n$$\\mathbf{v} = \\frac{10\\mathbf{i} - 24\\mathbf{j}}{2}$$\n\n$$\\mathbf{v} = 5\\mathbf{i} - 12\\mathbf{j}$$\n\nThus, the constant velocity of the toy boat $P$ is $(5\\mathbf{i} - 12\\mathbf{j}) \\text{ m s}^{-1}$."}]},{"id":"component_wise","label":"Component-by-Component Approach","steps":[{"type":"text","order":0,"title":"Analyze the components","content":"To solve this, we can treat the horizontal ($\\mathbf{i}$) and vertical ($\\mathbf{j}$) motions completely independently.\n\nThe constant velocity vector can be written as:\n$$\\mathbf{v} = v_x\\mathbf{i} + v_y\\mathbf{j}$$\n\nLet's write down the initial and final position components:\n- **Initially ($t = 0$)**: \n $$x_0 = -2 \\text{ m}, \\quad y_0 = 7 \\text{ m}$$\n- **At $t = 2$ seconds**:\n $$x_2 = 8 \\text{ m}, \\quad y_2 = -17 \\text{ m}$$\n\nOur goal is to find the velocity components $v_x$ and $v_y$."},{"type":"text","order":1,"title":"Calculate horizontal velocity component","content":"Let's look at the horizontal direction (the $\\mathbf{i}$ component).\n\nSince the velocity is constant, the horizontal displacement is given by:\n$$x_2 = x_0 + v_x t$$\n\nSubstitute the known values ($x_0 = -2$, $x_2 = 8$, and $t = 2$):\n$$8 = -2 + 2v_x$$\n\n$$10 = 2v_x$$\n\n$$v_x = 5 \\text{ m s}^{-1}$$"},{"type":"text","order":2,"title":"Calculate vertical velocity component","content":"Now let's look at the vertical direction (the $\\mathbf{j}$ component).\n\nSimilarly, the vertical displacement is given by:\n$$y_2 = y_0 + v_y t$$\n\nSubstitute the known values ($y_0 = 7$, $y_2 = -17$, and $t = 2$):\n$$-17 = 7 + 2v_y$$\n\n$$-24 = 2v_y$$\n\n$$v_y = -12 \\text{ m s}^{-1}$$"},{"type":"text","order":3,"title":"Combine the components","content":"Now we combine our horizontal and vertical velocity components back into vector form:\n$$\\mathbf{v} = v_x\\mathbf{i} + v_y\\mathbf{j}$$\n\nSubstituting $v_x = 5$ and $v_y = -12$:\n$$\\mathbf{v} = 5\\mathbf{i} - 12\\mathbf{j}$$"}]}],"generatedAt":"2026-06-09T04:05:39.777Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"76ab3fda-4ef6-4ec4-b758-b1fa0a110d8d","label":"b","content":"When $t = 3$ $P$ is at the point $C$. Find the distance $AC$.","metadata":null,"markscheme":"- Find position at $t=3$: $\\mathbf{r}_C = (-2\\mathbf{i} + 7\\mathbf{j}) + 3(5\\mathbf{i} - 12\\mathbf{j}) = 13\\mathbf{i} - 29\\mathbf{j}$ M1\n- Find displacement vector $\\vec{AC} = \\mathbf{r}_C - \\mathbf{r}_A = 15\\mathbf{i} - 36\\mathbf{j}$ M1\n- Calculate magnitude: $\\sqrt{15^2 + (-36)^2}$ M1\n- Final answer: $39 \\text{ m}$ A1","marks":4,"order":1,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Calculate the position vector of $C$ using $t=3$.","code":"M"},{"id":"p2","marks":1,"description":"Determine the displacement vector from $A$ to $C$.","code":"M"},{"id":"p3","marks":1,"description":"Use Pythagoras theorem to find distance.","code":"M"},{"id":"p4","marks":1,"description":"Final distance $39 \\text{ m}$.","code":"A"}],"finalAnswers":["39","39 \\text{ m}"],"generatedAt":"2026-05-06T14:03:50.806Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"position_vector","label":"Position Vector Method","steps":[{"type":"text","order":0,"title":"Find the constant velocity","content":"To find the velocity of the boat, we look at the displacement from point $A$ to point $B$ over the first $2$ seconds.\n\nThe initial position vector at $t = 0$ is:\n$$\\mathbf{r}_A = -2\\mathbf{i} + 7\\mathbf{j}$$\n\nAt $t = 2$, the position vector is:\n$$\\mathbf{r}_B = 8\\mathbf{i} - 17\\mathbf{j}$$\n\nThe constant velocity vector $\\mathbf{v}$ is given by:\n$$\\mathbf{v} = \\frac{\\mathbf{r}_B - \\mathbf{r}_A}{\\Delta t}$$\n\n$$\\mathbf{v} = \\frac{(8\\mathbf{i} - 17\\mathbf{j}) - (-2\\mathbf{i} + 7\\mathbf{j})}{2} = \\frac{10\\mathbf{i} - 24\\mathbf{j}}{2} = 5\\mathbf{i} - 12\\mathbf{j}$$","highlights":[{"text":"Initially $P$ is at the point $A$","location":"specification"},{"text":"At time $t = 2$ seconds, $P$ is at the point $B$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"We substitute t = 3 into the position vector formula to find the position of point C:"},{"latex":"$$\\mathbf{r}_C = \\htmlData{anchor=pos-init}{\\color{@sky}{-2\\mathbf{i} + 7\\mathbf{j}}} + \\htmlData{anchor=t-val}{\\color{@purple}{3}}\\htmlData{anchor=vel-val}{\\color{@amber}{(5\\mathbf{i} - 12\\mathbf{j})}}$"},{"latex":"$$\\mathbf{r}_C = (-2\\mathbf{i} + 7\\mathbf{j}) + (15\\mathbf{i} - 36\\mathbf{j}) = 13\\mathbf{i} - 29\\mathbf{j}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"pos-init"},{"color":"amber","style":"underline","anchor":"vel-val"},{"color":"purple","style":"underline","anchor":"t-val"}],"connections":[{"to":"pos-init","from":"vel-val","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Find position of C","description":"We calculate the position vector of the boat at $t = 3$ seconds."},{"type":"text","order":2,"title":"Find displacement vector AC","content":"Next, we find the displacement vector $\\vec{AC}$ by subtracting the position vector of $A$ from the position vector of $C$:\n\n$$\\vec{AC} = \\mathbf{r}_C - \\mathbf{r}_A$$\n\n$$\\vec{AC} = (13\\mathbf{i} - 29\\mathbf{j}) - (-2\\mathbf{i} + 7\\mathbf{j})$$\n\n$$\\vec{AC} = 15\\mathbf{i} - 36\\mathbf{j}$$"},{"type":"text","order":3,"title":"Calculate the magnitude","content":"The distance $AC$ is the magnitude of the displacement vector $\\vec{AC}$. We use the Pythagorean theorem to calculate this:\n\n$$AC = |\\vec{AC}| = \\sqrt{15^2 + (-36)^2}$$\n\n$$AC = \\sqrt{225 + 1296} = \\sqrt{1521} = 39\\text{ m}$$","calculator":[{"gdc":{"expression":"√(15^2 + (-36)^2)","instructions":"Enter the square root of the sum of the squares of the components."},"ti84":{"expression":"√(15² + (-36)²)","instructions":"Press [2nd] [x^2] (to get the square root), then enter 15^2 + (-36)^2 and press [ENTER]."},"desmos":{"expression":"\\sqrt{15^2 + (-36)^2}","instructions":"Type `sqrt(15^2 + (-36)^2)` in the expression bar."},"description":"Calculate the magnitude of the vector $15\\mathbf{i} - 36\\mathbf{j}$"}]}]},{"id":"vector_scaling","label":"Vector Scaling Method","steps":[{"type":"text","order":0,"title":"Calculate displacement vector AB","content":"Since the boat moves with constant velocity, we can use the position vectors at $t = 0$ (point $A$) and $t = 2$ (point $B$) to find the displacement vector $\\vec{AB}$ during this $2$-second interval.\n\nUsing the position vectors:\n$$\\mathbf{r}_A = -2\\mathbf{i} + 7\\mathbf{j}$$\n$$\\mathbf{r}_B = 8\\mathbf{i} - 17\\mathbf{j}$$\n\nWe find the displacement vector $\\vec{AB}$:\n$$\\vec{AB} = \\mathbf{r}_B - \\mathbf{r}_A$$\n$$\\vec{AB} = (8\\mathbf{i} - 17\\mathbf{j}) - (-2\\mathbf{i} + 7\\mathbf{j}) = 10\\mathbf{i} - 24\\mathbf{j}$$","highlights":[{"text":"Initially $P$ is at the point $A$","location":"specification"},{"text":"At time $t = 2$ seconds, $P$ is at the point $B$","location":"specification"}]},{"type":"text","order":1,"title":"Determine the scaling factor","content":"Because the boat's velocity is constant, its displacement is directly proportional to the elapsed time.\n\n- The displacement $\\vec{AB}$ occurs over a time interval of $t = 2$ seconds.\n- The displacement $\\vec{AC}$ occurs over a time interval of $t = 3$ seconds.\n\nTherefore, we can find the scaling factor to relate these two displacements:\n$$\\text{Scaling Factor} = \\frac{3\\text{ seconds}}{2\\text{ seconds}} = 1.5$$","highlights":[{"text":"When $t = 3$ $P$ is at the point $C$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\vec{AB} = \\htmlData{anchor=ab-vec}{\\color{@sky}{10\\mathbf{i} - 24\\mathbf{j}}}$"},{"latex":"$$\\vec{AC} = \\htmlData{anchor=factor}{\\color{@amber}{1.5}} \\vec{AB}$"},{"latex":"$$\\vec{AC} = \\color{@amber}{1.5}(\\color{@sky}{10\\mathbf{i} - 24\\mathbf{j}}) = \\htmlData{anchor=ac-vec}{\\color{@purple}{15\\mathbf{i} - 36\\mathbf{j}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"ab-vec"},{"color":"amber","style":"text-color","anchor":"factor"},{"color":"purple","style":"text-color","anchor":"ac-vec"}],"connections":[{"to":"ac-vec","from":"ab-vec","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Scale to find AC","description":"We multiply the displacement vector $\\vec{AB}$ by the scaling factor $1.5$ to find the total displacement vector $\\vec{AC}$."},{"type":"text","order":3,"title":"Find the distance AC","content":"The distance $AC$ is the magnitude of the displacement vector $\\vec{AC} = 15\\mathbf{i} - 36\\mathbf{j}$. We calculate this using the Pythagorean theorem:\n\n$$AC = |\\vec{AC}| = \\sqrt{15^2 + (-36)^2} = \\sqrt{1521} = 39\\text{ m}$$","highlights":[{"text":"Find the distance $AC$","location":"specification"}]}]},{"id":"distance_scaling","label":"Distance Scaling Method","steps":[{"type":"text","order":0,"title":"Calculate displacement vector AB","content":"To find the actual distance $AB$, we first calculate the displacement vector $\\vec{AB}$ from point $A$ to point $B$.\n\nThe position vector of $A$ (initially at $t = 0$) is:\n$$\\mathbf{r}_A = -2\\mathbf{i} + 7\\mathbf{j}$$\n\nThe position vector of $B$ (at $t = 2$) is:\n$$\\mathbf{r}_B = 8\\mathbf{i} - 17\\mathbf{j}$$\n\nSubtract $\\mathbf{r}_A$ from $\\mathbf{r}_B$ to find $\\vec{AB}$:\n$$\\vec{AB} = \\mathbf{r}_B - \\mathbf{r}_A$$\n$$\\vec{AB} = (8\\mathbf{i} - 17\\mathbf{j}) - (-2\\mathbf{i} + 7\\mathbf{j}) = 10\\mathbf{i} - 24\\mathbf{j}$$","highlights":[{"text":"Initially $P$ is at the point $A$","location":"specification"},{"text":"At time $t = 2$ seconds, $P$ is at the point $B$","location":"specification"}]},{"type":"text","order":1,"title":"Find the distance AB","content":"Now, we find the magnitude of the displacement vector $\\vec{AB}$ to get the distance between $A$ and $B$:\n\n$$|\\vec{AB}| = \\sqrt{10^2 + (-24)^2} = \\sqrt{676} = 26\\text{ m}$$"},{"type":"text","order":2,"title":"Calculate the constant speed","content":"The boat travels from $A$ to $B$ in $2$ seconds (from $t = 0$ to $t = 2$).\n\nSince the boat has constant velocity, its constant speed $v$ is the distance divided by the time taken:\n\n$$v = \\frac{\\text{Distance } AB}{\\text{Time}} = \\frac{26 \\text{ m}}{2 \\text{ s}} = 13 \\text{ m s}^{-1}$$"},{"type":"text","order":3,"title":"Scale to find distance AC","content":"We need to find the distance $AC$ at $t = 3$ seconds.\n\nSince the boat starts at $A$ (where $t = 0$), the travel time from $A$ to $C$ is exactly $3$ seconds.\n\nSince speed is constant, we can multiply the speed by this time interval directly to find the distance $AC$:\n\n$$\\text{Distance } AC = \\text{Speed} \\times \\text{Time} = 13 \\text{ m s}^{-1} \\times 3 \\text{ s} = 39 \\text{ m}$$","highlights":[{"text":"Find the distance $AC$","location":"specification"}]}]}],"generatedAt":"2026-06-09T04:07:33.497Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"492a1650-fea3-4b56-9c84-f2d49535af73:vector_equation","label":"(a) Vector Equation Approach","steps":[{"type":"text","order":0,"title":"Identify the vector equation","content":"For a particle moving with constant velocity, we can use the vector equation of a straight line:\n\n$$\\mathbf{r} = \\mathbf{r}_0 + \\mathbf{v}t$$\n\nwhere:\n- $\\mathbf{r}_0$ is the initial position vector.\n- $\\mathbf{v}$ is the constant velocity vector.\n- $\\mathbf{r}$ is the position vector at time $t$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\mathbf{r}_0 = \\htmlData{anchor=r0-val}{\\color{@sky}{-2\\mathbf{i} + 7\\mathbf{j}}}, \\; \\mathbf{r} = \\htmlData{anchor=r-val}{\\color{@amber}{8\\mathbf{i} - 17\\mathbf{j}}}, \\; t = \\htmlData{anchor=t-val}{\\color{@purple}{2}}$"},{"latex":"$$\\htmlData{anchor=r-eq}{\\color{@amber}{\\mathbf{r}}} = \\htmlData{anchor=r0-eq}{\\color{@sky}{\\mathbf{r}_0}} + \\mathbf{v}\\htmlData{anchor=t-eq}{\\color{@purple}{t}}$"},{"latex":"$$\\color{@amber}{8\\mathbf{i} - 17\\mathbf{j}} = (\\color{@sky}{-2\\mathbf{i} + 7\\mathbf{j}}) + 2\\mathbf{v}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"r0-val"},{"color":"amber","style":"text-color","anchor":"r-val"},{"color":"purple","style":"text-color","anchor":"t-val"}],"connections":[{"to":"r0-eq","from":"r0-val","color":"sky","label":"initial position","style":"solid"},{"to":"r-eq","from":"r-val","color":"amber","label":"final position","style":"solid"},{"to":"t-eq","from":"t-val","color":"purple","label":"time elapsed","style":"solid"}]},"order":1,"title":"Substitute the given values","description":"We substitute the given initial position vector, the final position vector, and the time into our equation."},{"type":"text","order":2,"title":"Isolate the velocity term","content":"To solve for $\\mathbf{v}$, we first subtract the initial position vector $(-2\\mathbf{i} + 7\\mathbf{j})$ from both sides to group the $\\mathbf{i}$ and $\\mathbf{j}$ components together:\n\n$$2\\mathbf{v} = (8\\mathbf{i} - 17\\mathbf{j}) - (-2\\mathbf{i} + 7\\mathbf{j})$$\n\n$$2\\mathbf{v} = (8 - (-2))\\mathbf{i} + (-17 - 7)\\mathbf{j}$$\n\n$$2\\mathbf{v} = 10\\mathbf{i} - 24\\mathbf{j}$$"},{"type":"text","order":3,"title":"Solve for the velocity vector","content":"Finally, we divide both sides of the equation by $2$ to isolate $\\mathbf{v}$:\n\n$$\\mathbf{v} = \\frac{10\\mathbf{i} - 24\\mathbf{j}}{2}$$\n\n$$\\mathbf{v} = 5\\mathbf{i} - 12\\mathbf{j}$$\n\nThus, the constant velocity of the toy boat $P$ is $(5\\mathbf{i} - 12\\mathbf{j}) \\text{ m s}^{-1}$."}]},{"id":"492a1650-fea3-4b56-9c84-f2d49535af73:component_wise","label":"(a) Component-by-Component Approach","steps":[{"type":"text","order":0,"title":"Analyze the components","content":"To solve this, we can treat the horizontal ($\\mathbf{i}$) and vertical ($\\mathbf{j}$) motions completely independently.\n\nThe constant velocity vector can be written as:\n$$\\mathbf{v} = v_x\\mathbf{i} + v_y\\mathbf{j}$$\n\nLet's write down the initial and final position components:\n- **Initially ($t = 0$)**: \n $$x_0 = -2 \\text{ m}, \\quad y_0 = 7 \\text{ m}$$\n- **At $t = 2$ seconds**:\n $$x_2 = 8 \\text{ m}, \\quad y_2 = -17 \\text{ m}$$\n\nOur goal is to find the velocity components $v_x$ and $v_y$."},{"type":"text","order":1,"title":"Calculate horizontal velocity component","content":"Let's look at the horizontal direction (the $\\mathbf{i}$ component).\n\nSince the velocity is constant, the horizontal displacement is given by:\n$$x_2 = x_0 + v_x t$$\n\nSubstitute the known values ($x_0 = -2$, $x_2 = 8$, and $t = 2$):\n$$8 = -2 + 2v_x$$\n\n$$10 = 2v_x$$\n\n$$v_x = 5 \\text{ m s}^{-1}$$"},{"type":"text","order":2,"title":"Calculate vertical velocity component","content":"Now let's look at the vertical direction (the $\\mathbf{j}$ component).\n\nSimilarly, the vertical displacement is given by:\n$$y_2 = y_0 + v_y t$$\n\nSubstitute the known values ($y_0 = 7$, $y_2 = -17$, and $t = 2$):\n$$-17 = 7 + 2v_y$$\n\n$$-24 = 2v_y$$\n\n$$v_y = -12 \\text{ m s}^{-1}$$"},{"type":"text","order":3,"title":"Combine the components","content":"Now we combine our horizontal and vertical velocity components back into vector form:\n$$\\mathbf{v} = v_x\\mathbf{i} + v_y\\mathbf{j}$$\n\nSubstituting $v_x = 5$ and $v_y = -12$:\n$$\\mathbf{v} = 5\\mathbf{i} - 12\\mathbf{j}$$"}]},{"id":"76ab3fda-4ef6-4ec4-b758-b1fa0a110d8d:position_vector","label":"(b) Position Vector Method","steps":[{"type":"text","order":0,"title":"Find the constant velocity","content":"To find the velocity of the boat, we look at the displacement from point $A$ to point $B$ over the first $2$ seconds.\n\nThe initial position vector at $t = 0$ is:\n$$\\mathbf{r}_A = -2\\mathbf{i} + 7\\mathbf{j}$$\n\nAt $t = 2$, the position vector is:\n$$\\mathbf{r}_B = 8\\mathbf{i} - 17\\mathbf{j}$$\n\nThe constant velocity vector $\\mathbf{v}$ is given by:\n$$\\mathbf{v} = \\frac{\\mathbf{r}_B - \\mathbf{r}_A}{\\Delta t}$$\n\n$$\\mathbf{v} = \\frac{(8\\mathbf{i} - 17\\mathbf{j}) - (-2\\mathbf{i} + 7\\mathbf{j})}{2} = \\frac{10\\mathbf{i} - 24\\mathbf{j}}{2} = 5\\mathbf{i} - 12\\mathbf{j}$$","highlights":[{"text":"Initially $P$ is at the point $A$","location":"specification"},{"text":"At time $t = 2$ seconds, $P$ is at the point $B$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"We substitute t = 3 into the position vector formula to find the position of point C:"},{"latex":"$$\\mathbf{r}_C = \\htmlData{anchor=pos-init}{\\color{@sky}{-2\\mathbf{i} + 7\\mathbf{j}}} + \\htmlData{anchor=t-val}{\\color{@purple}{3}}\\htmlData{anchor=vel-val}{\\color{@amber}{(5\\mathbf{i} - 12\\mathbf{j})}}$"},{"latex":"$$\\mathbf{r}_C = (-2\\mathbf{i} + 7\\mathbf{j}) + (15\\mathbf{i} - 36\\mathbf{j}) = 13\\mathbf{i} - 29\\mathbf{j}$"}],"highlights":[{"color":"sky","style":"underline","anchor":"pos-init"},{"color":"amber","style":"underline","anchor":"vel-val"},{"color":"purple","style":"underline","anchor":"t-val"}],"connections":[{"to":"pos-init","from":"vel-val","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Find position of C","description":"We calculate the position vector of the boat at $t = 3$ seconds."},{"type":"text","order":2,"title":"Find displacement vector AC","content":"Next, we find the displacement vector $\\vec{AC}$ by subtracting the position vector of $A$ from the position vector of $C$:\n\n$$\\vec{AC} = \\mathbf{r}_C - \\mathbf{r}_A$$\n\n$$\\vec{AC} = (13\\mathbf{i} - 29\\mathbf{j}) - (-2\\mathbf{i} + 7\\mathbf{j})$$\n\n$$\\vec{AC} = 15\\mathbf{i} - 36\\mathbf{j}$$"},{"type":"text","order":3,"title":"Calculate the magnitude","content":"The distance $AC$ is the magnitude of the displacement vector $\\vec{AC}$. We use the Pythagorean theorem to calculate this:\n\n$$AC = |\\vec{AC}| = \\sqrt{15^2 + (-36)^2}$$\n\n$$AC = \\sqrt{225 + 1296} = \\sqrt{1521} = 39\\text{ m}$$","calculator":[{"gdc":{"expression":"√(15^2 + (-36)^2)","instructions":"Enter the square root of the sum of the squares of the components."},"ti84":{"expression":"√(15² + (-36)²)","instructions":"Press [2nd] [x^2] (to get the square root), then enter 15^2 + (-36)^2 and press [ENTER]."},"desmos":{"expression":"\\sqrt{15^2 + (-36)^2}","instructions":"Type `sqrt(15^2 + (-36)^2)` in the expression bar."},"description":"Calculate the magnitude of the vector $15\\mathbf{i} - 36\\mathbf{j}$"}]}]},{"id":"76ab3fda-4ef6-4ec4-b758-b1fa0a110d8d:vector_scaling","label":"(b) Vector Scaling Method","steps":[{"type":"text","order":0,"title":"Calculate displacement vector AB","content":"Since the boat moves with constant velocity, we can use the position vectors at $t = 0$ (point $A$) and $t = 2$ (point $B$) to find the displacement vector $\\vec{AB}$ during this $2$-second interval.\n\nUsing the position vectors:\n$$\\mathbf{r}_A = -2\\mathbf{i} + 7\\mathbf{j}$$\n$$\\mathbf{r}_B = 8\\mathbf{i} - 17\\mathbf{j}$$\n\nWe find the displacement vector $\\vec{AB}$:\n$$\\vec{AB} = \\mathbf{r}_B - \\mathbf{r}_A$$\n$$\\vec{AB} = (8\\mathbf{i} - 17\\mathbf{j}) - (-2\\mathbf{i} + 7\\mathbf{j}) = 10\\mathbf{i} - 24\\mathbf{j}$$","highlights":[{"text":"Initially $P$ is at the point $A$","location":"specification"},{"text":"At time $t = 2$ seconds, $P$ is at the point $B$","location":"specification"}]},{"type":"text","order":1,"title":"Determine the scaling factor","content":"Because the boat's velocity is constant, its displacement is directly proportional to the elapsed time.\n\n- The displacement $\\vec{AB}$ occurs over a time interval of $t = 2$ seconds.\n- The displacement $\\vec{AC}$ occurs over a time interval of $t = 3$ seconds.\n\nTherefore, we can find the scaling factor to relate these two displacements:\n$$\\text{Scaling Factor} = \\frac{3\\text{ seconds}}{2\\text{ seconds}} = 1.5$$","highlights":[{"text":"When $t = 3$ $P$ is at the point $C$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\vec{AB} = \\htmlData{anchor=ab-vec}{\\color{@sky}{10\\mathbf{i} - 24\\mathbf{j}}}$"},{"latex":"$$\\vec{AC} = \\htmlData{anchor=factor}{\\color{@amber}{1.5}} \\vec{AB}$"},{"latex":"$$\\vec{AC} = \\color{@amber}{1.5}(\\color{@sky}{10\\mathbf{i} - 24\\mathbf{j}}) = \\htmlData{anchor=ac-vec}{\\color{@purple}{15\\mathbf{i} - 36\\mathbf{j}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"ab-vec"},{"color":"amber","style":"text-color","anchor":"factor"},{"color":"purple","style":"text-color","anchor":"ac-vec"}],"connections":[{"to":"ac-vec","from":"ab-vec","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Scale to find AC","description":"We multiply the displacement vector $\\vec{AB}$ by the scaling factor $1.5$ to find the total displacement vector $\\vec{AC}$."},{"type":"text","order":3,"title":"Find the distance AC","content":"The distance $AC$ is the magnitude of the displacement vector $\\vec{AC} = 15\\mathbf{i} - 36\\mathbf{j}$. We calculate this using the Pythagorean theorem:\n\n$$AC = |\\vec{AC}| = \\sqrt{15^2 + (-36)^2} = \\sqrt{1521} = 39\\text{ m}$$","highlights":[{"text":"Find the distance $AC$","location":"specification"}]}]},{"id":"76ab3fda-4ef6-4ec4-b758-b1fa0a110d8d:distance_scaling","label":"(b) Distance Scaling Method","steps":[{"type":"text","order":0,"title":"Calculate displacement vector AB","content":"To find the actual distance $AB$, we first calculate the displacement vector $\\vec{AB}$ from point $A$ to point $B$.\n\nThe position vector of $A$ (initially at $t = 0$) is:\n$$\\mathbf{r}_A = -2\\mathbf{i} + 7\\mathbf{j}$$\n\nThe position vector of $B$ (at $t = 2$) is:\n$$\\mathbf{r}_B = 8\\mathbf{i} - 17\\mathbf{j}$$\n\nSubtract $\\mathbf{r}_A$ from $\\mathbf{r}_B$ to find $\\vec{AB}$:\n$$\\vec{AB} = \\mathbf{r}_B - \\mathbf{r}_A$$\n$$\\vec{AB} = (8\\mathbf{i} - 17\\mathbf{j}) - (-2\\mathbf{i} + 7\\mathbf{j}) = 10\\mathbf{i} - 24\\mathbf{j}$$","highlights":[{"text":"Initially $P$ is at the point $A$","location":"specification"},{"text":"At time $t = 2$ seconds, $P$ is at the point $B$","location":"specification"}]},{"type":"text","order":1,"title":"Find the distance AB","content":"Now, we find the magnitude of the displacement vector $\\vec{AB}$ to get the distance between $A$ and $B$:\n\n$$|\\vec{AB}| = \\sqrt{10^2 + (-24)^2} = \\sqrt{676} = 26\\text{ m}$$"},{"type":"text","order":2,"title":"Calculate the constant speed","content":"The boat travels from $A$ to $B$ in $2$ seconds (from $t = 0$ to $t = 2$).\n\nSince the boat has constant velocity, its constant speed $v$ is the distance divided by the time taken:\n\n$$v = \\frac{\\text{Distance } AB}{\\text{Time}} = \\frac{26 \\text{ m}}{2 \\text{ s}} = 13 \\text{ m s}^{-1}$$"},{"type":"text","order":3,"title":"Scale to find distance AC","content":"We need to find the distance $AC$ at $t = 3$ seconds.\n\nSince the boat starts at $A$ (where $t = 0$), the travel time from $A$ to $C$ is exactly $3$ seconds.\n\nSince speed is constant, we can multiply the speed by this time interval directly to find the distance $AC$:\n\n$$\\text{Distance } AC = \\text{Speed} \\times \\text{Time} = 13 \\text{ m s}^{-1} \\times 3 \\text{ s} = 39 \\text{ m}$$","highlights":[{"text":"Find the distance 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3 Mechanics

Question 33

3 Mechanics Questions

3 Mechanics

Question 33

3 Mechanics Questions