![](\"resources/question-resources/Car_image.jpg\"/)
Its speed at $A$ is $\\var{u} \\mathrm{ms^{-1}}$ and it takes $\\var{t}$ seconds to move from $A$ to $B$.For 'Equations of motions' page on the wiki, under kinematics. Includes questions using the SUVAT equations.
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\n$a = \\var{a},$ $u=\\var{u},$ $t=\\var{t},$ $v=$ $?,$ $s=$ $?$
\na) At $B$ the car will have reached its final velocity, $v \\mathrm{ms^{-1}}$. You need $v$ and you know $u, a$ and $t$ so you can use the equation $v= u +at$. \\begin{align} v &= u + at, \\\\
&= \\var{u} + (\\var{a} \\times \\var{t}), \\\\
&= \\var{u +a*t} \\mathrm{ms^{-1}}. \\end{align}
The speed of the car at $B$ is $\\var{u + a*t} \\mathrm{ms^{-1}}.$
\nb) You need the distance, $s \\mathrm{m}$. You calculated $v$ in the previous part, so you can use the equation $s=\\left(\\frac{u+v}{2}\\right)t.$
\n\\begin{align} s &= \\left(\\frac{u+v}{2}\\right)t, \\\\
&= \\left(\\frac{\\var{u}+\\var{v}}{2}\\right) \\times \\var{t}, \\\\
&= \\var{((u+v)/2)*t} \\mathrm{m}. \\end{align}
The distance from $A$ to $B$ is $\\var{((u+v)/2)*t}$ metres.
", "rulesets": {}, "parts": [{"integerPartialCredit": 0, "prompt": "
Find the speed of the car in $\\mathrm{ms^{-1}}$ at $B$.
", "integerAnswer": true, "allowFractions": false, "variableReplacements": [], "maxValue": "u+a*t", "minValue": "u+a*t", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"prompt": "Find the distance in $\\mathrm{m}$ from $A$ to $B$.
", "allowFractions": false, "variableReplacements": [], "maxValue": "(u+v)*(0.5)*t", "minValue": "(u+v)*(0.5)*t", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "\nA car is driving in a straight line from $A$ to $B$ with constant acceleration $\\var{a} \\mathrm{ms^{-2}}$.
\nIts speed at $A$ is $\\var{u} \\mathrm{ms^{-1}}$ and it takes $\\var{t}$ seconds to move from $A$ to $B$.
constant acceleration
"}, "u": {"definition": "random(1..9#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "u", "description": "initial speed
"}, "t": {"definition": "random(1..9#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "t", "description": "time (seconds)
"}, "v": {"definition": "u+a*t", "templateType": "anything", "group": "Ungrouped variables", "name": "v", "description": "part a) answer. final velocity
"}}, "metadata": {"description": "SUVAT equation questions, for mechanics page in wiki. Uses $v=u+at$ and $s=((u+v)/2)t$.
", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equations question 2", "extensions": [], "custom_part_types": [], "resources": [["question-resources/truck_image.png", "/srv/numbas/media/question-resources/truck_image.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["u", "v", "t"], "tags": [], "advice": "You start by writing down the values you know and the values you need to find.
\n$a =$ $?$, $u=\\var{u},$ $t=\\var{t},$ $v=\\var{v},$ $s=$ $?$
\na) You need the distance $s \\mathrm{m}$ and you know $u, v$ and $t$ so you can use the equation $s= \\left(\\frac{u+v}{2}\\right)t$. \\begin{align} s &= \\left(\\frac{u+v}{2}\\right)t, \\\\
&= \\left(\\frac{\\var{u}+\\var{v}}{2}\\right) \\times \\var{t}, \\\\
&= \\var{(u+v)*0.5*t} \\mathrm{m}.\\end{align}
The distance the truck travels is $\\var{(u+v)*0.5*t}$ metres.
\nb) You need the acceleration, $a \\mathrm{ms^{-2}}$, and you know $u$,$v$ and $t$ so you can use the equation $v=u + at$ rearranged for $a$.
\n\\begin{align} a &= \\left(\\frac{v-u}{t}\\right), \\\\
&= \\left(\\frac{\\var{v}-\\var{u}}{\\var{t}}\\right), \\\\
&= \\frac{\\var{(v-u)}}{\\var{t}} \\mathrm{ms^{-2}}. \\end{align}
The acceleration of the truck is $\\frac{\\var{(v-u)}}{\\var{t}} \\mathrm{ms^{-2}}$.
Find the distance in $\\mathrm{m}$ that the truck travels in these $\\var{t}$ seconds.
", "allowFractions": false, "variableReplacements": [], "maxValue": "(u+v)*0.5*t", "minValue": "(u+v)*0.5*t", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"prompt": "Find the acceleration in $\\mathrm{ms^{-2}}$ of the truck in these $\\var{t}$ seconds.
", "allowFractions": true, "variableReplacements": [], "maxValue": "(v-u)/t", "minValue": "(v-u)/t", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "A truck accelerates along a straight road at a constant rate from a speed of $\\var{u} \\mathrm{ms^{-1}}$ to $\\var{v} \\mathrm{ms^{-1}}$ in $\\var{t}$ seconds.
\n![](\"resources/question-resources/truck_image.png\"/)
initial velocity
"}, "t": {"definition": "random(2..15#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "t", "description": "time
"}, "v": {"definition": "random(7..15#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "v", "description": "final velocity
"}}, "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equations question 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "A cyclist rides in a straight line with constant decelaration $\\var{-{a}} \\mathrm{ms^{-2}}$ from a point $A$ to a point $B$. At point $A$ the cyclist's speed is $\\var{u}\\mathrm{ms^{-1}}$ and at point $B$ the speed is $\\var{v}\\mathrm{ms^{-1}}$.
", "advice": "a) You are given $u = \\var{u},$ $v=\\var{v}$, $a = \\var{a}$, $t = ?$, $s = ?$
\nNote that $a$ is negative because the cyclist is decelerating.
\nYou need $t$, so you use $v=u+at$, rearranged for $t$.
\n\\begin{align} t & = \\frac{v-u}{a}, \\\\
& = \\frac{\\var{v}-\\var{u}}{\\var{a}}, \\\\
& = \\var[fractionNumbers]{(v-u)/a} \\mathrm{s}. \\end{align}
The time taken to move from $A$ to $B$ is $\\var[fractionNumbers]{(v-u)/a}$ seconds.
\nb) You need the distance, $s \\mathrm{m}$. Therefore you can use $s = \\left(\\frac{u+v}{2}\\right)t$, with the $t$ value found in part a).
\n\\begin{align} s & = \\left(\\frac{u+v}{2}\\right)t, \\\\
& = \\left(\\frac{\\var{u}+\\var{v}}{2}\\right) \\times \\var[fractionNumbers]{(v-u)/a}, \\\\
& = \\var[fractionNumbers]{(u+v)*0.5*((v-u)/a)} \\mathrm{m}. \\end{align}
The distance from $A$ to $B$ is $\\var[fractionNumbers]{(u+v)*0.5*((v-u)/a)}$ metres.
c) You now have $u=\\var{u}$, $a = \\var{a}$, $t=\\var{t2}$, $s=?$ and you are trying to find the velocity at point $C$ so $v=?$
\nTherefore you can use $v = u+at$.
\n\\begin{align} v & = u + at, \\\\
& = \\var{u} + \\left(\\var{a} \\times \\var{t2}\\right), \\\\
& = \\var[fractionNumbers]{u+a*t2} \\mathrm{ms^{-1}}. \\end{align}
Note that the velocity can be negative - this means the cyclist is moving from right to left. Therefore the velocity of the cyclist at point $C$ is $\\var{-(u+a*t2)} \\mathrm{ms^{-1}}$ in the direction $\\simplify{vec:BA}.$
\nd) We now need the distance, $s \\mathrm{m}$. We have $u=\\var{u}$, $t=\\var{t2}$ and we have just evaluated the final velocity at point C, $v=\\var{u+a*t2}$.
\n\\begin{align} s & = \\left(\\frac{u + v}{2}\\right)t, \\\\
& = \\left(\\frac{\\var{u} + \\var{u+a*t2}}{2}\\right) \\times \\var{t2}, \\\\
& = \\var[fractionNumbers]{(u+u+a*t2)*0.5*t2} \\mathrm{m}. \\end{align}
The distance from A to C is $\\var[fractionNumbers]{(u+u+a*t2)*0.5*t2}$ metres.
\n", "rulesets": {}, "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "random(-1.5 .. -0.5#0.25)", "description": "acceleration negative so that deceleration is positive
", "templateType": "randrange"}, "t2": {"name": "t2", "group": "Ungrouped variables", "definition": "random(10 .. 15#1)", "description": "", "templateType": "randrange"}, "s": {"name": "s", "group": "Ungrouped variables", "definition": "(u + v2)*0.5*t2\n", "description": "distance A to C
\n", "templateType": "anything"}, "v2": {"name": "v2", "group": "Ungrouped variables", "definition": "u+a*t2", "description": "
velocity at point C. can be negative
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", "templateType": "randrange"}}, "variablesTest": {"condition": "[s>0,v2<0,t2>t,abs(v2)Find the distance in $\\mathrm{m}$ from $A$ to $B$.
", "minValue": "(u+v)*0.5*t", "maxValue": "(u+v)*0.5*t", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Suppose that, after reaching $B$ the cyclist continues to move along the same straight line with constant deceleration $\\var{-a} \\mathrm{ms^{-2}}$. The cyclist is then at a point $C$ $\\var{t2}$ seconds after they leave point $A$. Find the velocity in $\\mathrm{ms^{-1}}$ of the cyclist at point $C$.
", "minValue": "u+a*t2", "maxValue": "u+a*t2", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Find the distance in $\\mathrm{m}$ from $A$ to $C$.
", "minValue": "(u+v2)*0.5*t2", "maxValue": "(u+v2)*0.5*t2", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "SUVAT equations question 4", "extensions": [], "custom_part_types": [], "resources": [["question-resources/man_image.png", "/srv/numbas/media/question-resources/man_image.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["t", "km"], "tags": [], "advice": "a) It is important to always convert your measurements to base SI units before using the SUVAT equations. For velocity, the SI units are $\\mathrm{ms^{-1}}$. We have $\\mathrm{km \\ h^{-1}}$, where $1\\mathrm{km} = 1000\\mathrm{m}$ and $1\\mathrm{h} = 60 \\times 60 \\mathrm{s} = 3600\\mathrm{s}$.
\nTherefore
\n\\begin{align} \\var{km} \\mathrm{km \\ h^{-1}} & = \\var{km} \\times 1000 \\div 3600 \\mathrm{ms^{-1}},\\\\ & = \\var[fractionNumbers]{km*1000/3600}\\mathrm{ms^{-1}}. \\end{align}
\nb) As the pedestrian starts from rest $u=0$. Then we have $v=\\var[fractionNumbers]{km*1000/3600}$, $t=\\var{t}$, $a=?$ and $s=?$. We want $a$ so we can use $v=u+at$ rearranged for $a$, remembering to use our $v$ in SI units from part a).
\n\\begin{align} a &= \\frac{v-u}{t}, \\\\
&= \\frac{\\var[fractionNumbers]{km*1000/3600} - 0}{\\var{t}}, \\\\
&= \\var[fractionNumbers]{(km*1000/3600)/t} \\mathrm{ms^{-2}}. \\end{align}
The acceleration of the pedestrian is $\\var[fractionNumbers]{(km*1000/3600)/t} \\mathrm{ms^{-2}}.$
\nc) We want the distance, $s$. So we use $s = \\left(\\frac{u+v}{2}\\right)t$.
\n\\begin{align} s & = \\left(\\frac{u+v}{2}\\right)t, \\\\
& = \\left(\\frac{0+\\var[fractionNumbers]{km*1000/3600}}{2}\\right)\\times \\var{t}, \\\\
& = \\var[fractionNumbers]{(km*1000/3600)*0.5*t} \\mathrm{m}.\\end{align}
The distance between the crossing and the shop is $\\var[fractionNumbers]{(km*1000/3600)*0.5*t}\\mathrm{m}.$
\n", "rulesets": {}, "parts": [{"integerPartialCredit": 0, "prompt": "
Convert $\\var{km} \\mathrm{km \\ h^{-1}}$ into SI units $\\mathrm{ms^{-1}}$ (enter your answer as a fraction).
", "integerAnswer": true, "allowFractions": true, "variableReplacements": [], "maxValue": "km*1000/3600", "minValue": "km*1000/3600", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"prompt": "What is the pedestrain's accelerationin $\\mathrm{ms^{-2}}$?
", "allowFractions": true, "variableReplacements": [], "maxValue": "(km*1000/3600)/t", "minValue": "(km*1000/3600)/t", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"prompt": "Find the distance in $\\mathrm{m}$ from the crossing to the shop.
", "allowFractions": true, "variableReplacements": [], "maxValue": "(km*1000/3600)*0.5*t", "minValue": "(km*1000/3600)*0.5*t", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "A pedestrian waits at a crossing until the road is clear. He crosses the road with constant acceleration, starting from rest and $\\var{t}$ seconds later he passes a shop and is travelling at $\\var{km} \\mathrm{km \\ h^{-1}}.$
\n![](\"resources/question-resources/man_image.png\"/)
time he passes the shop
"}, "km": {"definition": "random(1..8#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "km", "description": "pedestrains speed in km. students convert to metres.
"}}, "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equations question 5", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["a", "u", "v"], "tags": [], "advice": "a) We have that $a=\\var{a}$, $u=\\var{u}$, $v=\\var{v}$. We want to find the distance, $s \\mathrm{m}$. We can use the formula $v^2=u^2+2as$ rearranged for $s$.
\nTherefore
\n\\begin{align} s & = \\frac{v^2 - u^2}{2a}, \\\\
& = \\frac{\\var{v}^2 - \\var{u}^2}{2\\times \\var{a}}, \\\\
& = \\var[fractionNumbers]{(v^2 - u^2)/(2*a)} \\mathrm{m}. \\end{align}
So the distance from $A$ to $B$ is $\\var[fractionNumbers]{(v^2 - u^2)/(2*a)} \\mathrm{m}$.
\nb) We have all the variables now except for $t$. To find the time travelled we can use the majority of the SUVAT equations, however it is best to use $v=u+at$ rearranged for $t$ as we are given $v, u$ and $a$ in the question, just incase the $s$ that was calculated previously was incorrect.
\nTherefore
\n\\begin{align} t &= \\frac{v-u}{a}, \\\\
&= \\frac{\\var{v}-\\var{u}}{\\var{a}}, \\\\
&= \\var[fractionNumbers]{(v-u)/a} \\mathrm{s}. \\end{align}
The time taken for the car to travel from $A$ to $B$ is $\\var[fractionNumbers]{(v-u)/a} \\mathrm{s}$.
", "rulesets": {}, "parts": [{"prompt": "Find the distance in $\\mathrm{m}$ from $A$ to $B$.
", "allowFractions": true, "variableReplacements": [], "maxValue": "(v^2-u^2)/(2*a)", "minValue": "(v^2-u^2)/(2*a)", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"prompt": "Find the time taken in seconds to travel from $A$ to $B$.
", "allowFractions": true, "variableReplacements": [], "maxValue": "(v-u)/a", "minValue": "(v-u)/a", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "A car is driving alone a straight road from $A$ to $B$ with constant acceleration $\\var{a} \\mathrm{ms^{-2}}$. At point $A$ the car is travelling at $\\var{u} \\mathrm{ms^{-1}}$ in the direction $\\simplify{vec:AB}$. At point $B$ the car is travelling in the same direction with velocity $\\var{v} \\mathrm{ms^{-1}}$.
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": "v>u"}, "preamble": {"css": "", "js": ""}, "variables": {"a": {"definition": "random(1..6#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "a", "description": "acceleration
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"}, "v": {"definition": "random(6..15#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "v", "description": "final velocity at point B
"}}, "metadata": {"description": "Section 2.2 of M1 book. Using v squared formula
", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equations question 6", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["a", "u", "t"], "tags": [], "advice": "a) We are given $a=\\var{a}, u=\\var{u}$ and $t=\\var{t}$. We are asked to find the distance, $s \\mathrm{m}$ therefore we can use the formula $s=ut+\\frac{1}{2}at^2.$
\n\\begin{align} s &= ut+\\frac{1}{2}at^2, \\\\
&= \\var{u}\\times\\var{t} + \\frac{1}{2} \\times \\var{a} \\times \\var{t}^2, \\\\
&=\\var[fractionNumbers]{u*t+0.5*a*t^2} \\mathrm{m}. \\end{align}
So the total distance the particle has travelled is $\\var[fractionNumbers]{u*t+0.5*a*t^2} \\mathrm{m}.$
b) It is a good idea to use data given in the question, therefore to find $v$ we will use the formula $v=u+at$.
\n\\begin{align} v &= u +at, \\\\
&= \\var{u} + \\var{a} \\times \\var{t}, \\\\
&= \\var[fractionNumbers]{u+a*t} \\mathrm{ms^{-1}}. \\end{align}
The particle is travelling at $\\var[fractionNumbers]{u+a*t} \\mathrm{ms^{-1}}$ when it passes point $X$.
", "rulesets": {}, "parts": [{"prompt": "What is the total distance in $\\mathrm{m}$ the particle has travelled when it reaches point $X$?
", "allowFractions": true, "variableReplacements": [], "maxValue": "u*t+0.5*a*t^2", "minValue": "u*t+0.5*a*t^2", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"prompt": "At what speed in $\\mathrm{ms^{-1}}$ does the particle pass point $X$?
", "allowFractions": true, "variableReplacements": [], "maxValue": "u+a*t", "minValue": "u+a*t", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "A particle is moving alone a straight line with constant acceleration of $\\var{a} \\mathrm{ms^{-2}}$. Initially the particle travels at $\\var{u}\\mathrm{ms^{-1}}$, then $\\var{t}$ seconds later it hits a point $X$.
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"css": "", "js": ""}, "variables": {"a": {"definition": "random(1..5#0.5)", "templateType": "randrange", "group": "Ungrouped variables", "name": "a", "description": "acceleration
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"}}, "metadata": {"description": "uses $s=ut+\\frac{1}{2}at^2$
", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equations question 7", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["a", "s", "u", "v", "t"], "tags": [], "preamble": {"css": "", "js": ""}, "advice": "a) We have that $a=\\var{a}, u=\\var{u}$ and $s=\\var{s}$. Time is unknown so we use a formula without $t$, this is $v^2=u^2+2as$.
\nTherefore
\n\\begin{align} v &= \\sqrt{u^2+2as}, \\\\
&= \\sqrt{\\var{u}^2+\\left(2 \\times \\var{a} \\times \\var{s}\\right)}, \\\\
&= \\sqrt{\\var[fractionNumbers]{u^2+2*a*s}}\\mathrm{ms^{-1}}. \\end{align}
So the speed when she reaches the shop is $\\sqrt{\\var[fractionNumbers]{u^2+2*a*s}} \\mathrm{ms^{-1}}$.
\nb) We have been given $a,u$ and $s$. Therefore to calculate time, $t$, we should use the equation $s=ut+\\frac{1}{2}at^2$ rearranged for $t$. We could then check our answer using $v=u+at$ rearranged for $t$, with the value of $v$ we found in part a), however if we have made a mistake in the previous part this will affect our answer.
\nTherefore we have $\\frac{1}{2}at^2 + ut - s = 0$, so to solve for $t$ we can use the quadratic formula $\\frac{-b+\\sqrt{b^2-4ac}}{2a}$, with $b=u, a=\\frac{1}{2}a$ and $c=-s$. Note that we only take the positive root as we are trying to find time.
\nThis becomes
\n$t = \\frac{-u + \\sqrt{u^2 - \\left(4 \\times \\frac{1}{2}a \\times -s\\right)}}{2 \\times \\frac{1}{2}a}.$
\nTherefore
\n$t = \\frac{-\\var{u} + \\sqrt{\\var{u}^2 + \\left(2\\times \\var{a} \\times \\var{s}\\right)}}{\\var{a}} = \\var{precround((-u+sqrt(u^2+2a*s))/a,3)} \\mathrm{s}$ which has been evaluated to 3d.p.
\nSo the time taken to reach the shop is $\\var{precround((-u+sqrt(u^2+2a*s))/a,3)} \\mathrm{s}.$
\nWe can check this using
\n\\begin{align} v & = u + at, \\\\
& = \\var{u} + \\var{a} \\times \\var{precround((-u+sqrt(u^2+2a*s))/a,3)}, \\\\
& = \\var{u + a*precround((-u+sqrt(u^2+2a*s))/a,3)}, \\\\
& = \\sqrt{\\var[fractionNumbers]{u^2+2*a*s}}\\mathrm{ms^{-1}}. \\end{align}
which is the speed we found in part a).
", "rulesets": {}, "parts": [{"vsetrangepoints": 5, "prompt": "
What is the speed in $\\mathrm{ms^{-1}}$ of the cyclist when she reaches the shop? (You can input a square root using the sqrt() function, for example sqrt(2) for $\\sqrt{2}$)
How long in seconds does it take the cyclist to reach the shop? (Answer to 3d.p.)
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "maxValue": "(-u+(u^2+2*a*s)^(1/2))/a", "strictPrecision": true, "minValue": "(-u+(u^2+2*a*s)^(1/2))/a", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "A cyclist travels from her home to the shop with constant acceleration $\\var{a} \\mathrm{ms^{-2}}$. The shop is $\\var{s} \\mathrm{m}$ away from her home and she initially travels at $\\var{u} \\mathrm{ms^{-1}}$.
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": "t>0"}, "variables": {"a": {"definition": "random(0.5..4#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "a", "description": "acceleration
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\nWe are trying to find the time $t$, so we use the formula $s=ut+\\frac{1}{2}at^2$. As this is a quadratic it will have two solutions, so we will solve for $t$ then input the smallest value as our answer for part a) and the other for part b).
\nThe formula becomes $\\frac{1}{2}at^2 + ut - s = 0$, from this we can use the quadratic formula to solve for the two values of $t$.
\na) $ t = \\frac{-u + \\sqrt{u^2 + 2as}}{a} = \\frac{-\\var{u} + \\sqrt{\\var{u}^2 +\\left(2\\times \\var{a} \\times \\var{s}\\right)}}{\\var{a}} = \\var{precround(t2,3)} \\mathrm{s}.$
\nThe shortest time it takes for the ball to pass through point $B$ is $\\var{precround(t2,3)} \\mathrm{s}$.
\nb) $ t = \\frac{-u - \\sqrt{u^2 + 2as}}{a} = \\frac{-\\var{u} - \\sqrt{\\var{u}^2 +\\left(2\\times \\var{a} \\times \\var{s}\\right)}}{\\var{a}} = \\var{precround(t1,3)} \\mathrm{s}.$
\nThe other time the ball passes through point $B$ is $\\var{precround(t1,3)} \\mathrm{s}$.
\nc) We now need the velocity $v$, and as we have two values for $t$ we can use the formula $v= u +at$ with both these values to calculate our two velocities. Using $t=\\var{precround(t2,3)}$ we get
\\begin{align} v &= u + at, \\\\
&=\\var{u} + \\left(\\var{a}\\times\\var{precround(t2,3)}\\right), \\\\
&= \\var{precround(u+a*precround(t2,3),3)} \\mathrm{ms^{-1}}. \\end{align}
This is positive, therefore the ball is travelling at $\\var{precround(u+a*precround(t2,3),3)} \\mathrm{ms^{-1}}$ in the direction $\\simplify{vec:AB}$.
\nd) Following the same method as part c) only now using $t=\\var{precround(t1,3)}$ we get
\\begin{align} v &= u + at, \\\\
&=\\var{u} + \\left(\\var{a}\\times\\var{precround(t1,3)}\\right), \\\\
&= \\var{precround(u+a*precround(t1,3),3)} \\mathrm{ms^{-1}}. \\end{align}
This is negative, therefore the ball is travelling at $\\var{-(precround(u+a*precround(t1,3),3))} \\mathrm{ms^{-1}}$ in the direction $\\simplify{vec:BA}$.
\ne) The ball returns to point $A$ when $s=0$, so the displacement from the initial position is $0 \\mathrm{m}$.
We also have $u=\\var{u}$ and $a=\\var{a}$ and we are trying to find the time, $t$. Therefore we can use the formula $s=ut+\\frac{1}{2}at^2$ with $s=0$.
This becomes \\begin{align} ut+\\frac{1}{2}at^2 &= 0, \\\\
t(u + \\frac{1}{2}at) &= 0.
\\end{align}
We need to discard the solution $t=0$ (which represents the starting position of the ball) so we have $u + \\frac{1}{2}at=0$. Solving for $t$ this gives
\\begin{align} t & = - \\frac{2}{a}u, \\\\
&= \\frac{2}{\\var{-a}}\\times \\var{u}, \\\\
& = \\var{precround(-(2/a)*u,3)}\\mathrm{s}. \\end{align}
So the time taken for the ball to return to point $A$ is $\\var{precround(-(2/a)*u,3)} \\mathrm{s}$.
\n", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "Find the shortest time in $\\mathrm{s}$ (to 3d.p.) when the ball rolls through $B$. This will be in the direction $\\simplify{vec:AB}$.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "t2", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "t2", "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "Find the longest time in $\\mathrm{s}$ (to 3d.p.) when the ball rolls through $B$. This will be in the direction $\\simplify{vec:BA}$.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "t1", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 0, "minValue": "t1", "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "Find the velocity in $\\mathrm{ms^{-1}}$ (to 3d.p.) of the ball when it rolls through $B$ in the direction $\\simplify{vec:AB}$.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "u+a*precround(t2,3)", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 0, "minValue": "u+a*precround(t2,3)", "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "Find the velocity in $\\mathrm{ms^{-1}}$ (to 3d.p.) of the ball when it rolls through $B$ in the direction $\\simplify{vec:BA}$.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "u+a*precround(t1,3)", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "u+a*precround(t1,3)", "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "Find the time in $\\mathrm{s}$ (to 3d.p.) other than $t=0$ when the ball returns to point $A$.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "(-2*u)/a", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "(-2*u)/a", "type": "numberentry", "showPrecisionHint": false}], "statement": "A ball rolls in a straight horizontal line with constant decelaration $\\var{-a} \\mathrm{ms^{-2}}.$ At time $t=0$ the ball rolls past a point $A$ with speed $\\var{u} \\mathrm{ms^{-1}}$, travelling towards a point $B$ that is $\\var{s} \\mathrm{m}$ away, before returning in the direction of point $A$.
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": "[u^2+2*a*s>0, t1>t2, -u+(u^2+2*a*s)^(1/2)<0, t1>0, t2>0]"}, "preamble": {"css": "", "js": ""}, "variables": {"a": {"definition": "random(-5..-0.5#0.5)", "templateType": "randrange", "group": "Ungrouped variables", "name": "a", "description": "acceleration - negative so deceleration is positive
"}, "t2": {"definition": "(-u+(u^2+2*a*s)^(1/2))/a", "templateType": "anything", "group": "Ungrouped variables", "name": "t2", "description": ""}, "v2": {"definition": "u+a*t1", "templateType": "anything", "group": "Ungrouped variables", "name": "v2", "description": "velocity moves in opposite direction
"}, "t1": {"definition": "(-u-(u^2+2*a*s)^(1/2))/a", "templateType": "anything", "group": "Ungrouped variables", "name": "t1", "description": "largest time
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"}, "s": {"definition": "random(3..10#0.5)", "templateType": "randrange", "group": "Ungrouped variables", "name": "s", "description": "distance between A and B
"}, "u": {"definition": "random(13..25#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "u", "description": "initial velocity at point A
"}}, "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equations question 9", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["s", "v", "t", "a", "u"], "tags": [], "advice": "a) We have that $t=\\var{t}, s=\\var{s}$ and $v=\\var{v}$. We are asked to find the acceleration of the car, $a$, therefore we can use the formula $s=vt-\\frac{1}{2}at^2$, rearranged for $a$.
\nThis gives \\begin{align} s &= vt - \\frac{1}{2}at^2, \\\\
\\frac{1}{2}at^2 &= vt - s ,\\\\
a & = \\frac{2(vt-s)}{t^2}, \\\\
& = \\frac{2(\\var{v} \\times \\var{t} - \\var{s})}{\\var{t}^2}, \\\\
& = \\var{precround((2(v*t-s))/(t^2),3)}\\mathrm{m^{-2}}. \\end{align}
So the acceleration of the car is $\\var{precround((2(v*t-s))/(t^2),3)} \\mathrm{ms^{-2}}$.
\nb) We are asked to find the initial velocity, $u$. We could use the formula $v=u+at$, however it is best to work with the data given in the question in case our answer to part a) is incorrect. Therefore we will use $s=\\left(\\frac{u+v}{2}\\right)t$ rearranged for $u$.
\nThis gives \\begin{align} s & = \\left(\\frac{u+v}{2}\\right)t, \\\\
u & = \\frac{2s}{t} - v, \\\\
& = \\frac{2 \\times \\var{s}}{\\var{t}} - \\var{v}, \\\\
& = \\var{precround((2*s)/t-v,3)} \\mathrm{ms^{-1}}. \\end{align}
The car was travelling at $\\var{precround((2*s)/t-v,3)} \\mathrm{ms^{-1}}$ when it passed the traffic lights.
Find the acceleration in $\\mathrm{ms^{-2}}$ (to 3d.p.) of the car.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "a", "minValue": "a", "variableReplacementStrategy": "originalfirst", "strictPrecision": true, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "Find the speed in $\\mathrm{ms^{-1}}$ (to 3d.p.) with which the car passed the traffic lights.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "u", "minValue": "u", "variableReplacementStrategy": "originalfirst", "strictPrecision": true, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "A car is driving along a straight road with constant acceleration. The car first passes a set of traffic lights then $\\var{s} \\mathrm{m}$ later, travelling at $\\var{v} \\mathrm{ms^{-1}}$ it passes a school. The time it takes the car to go from the traffic lights to the school is $\\var{t}$ seconds.
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": "a>0\n"}, "preamble": {"css": "", "js": ""}, "variables": {"a": {"definition": "2*(v*t-s)/t^2", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": "acceleration
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"}, "u": {"definition": "(2*s)/t-v", "templateType": "anything", "group": "Ungrouped variables", "name": "u", "description": ""}, "t": {"definition": "random(6..9#0.5)", "templateType": "randrange", "group": "Ungrouped variables", "name": "t", "description": "time (seconds)
"}, "v": {"definition": "random(5..8#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "v", "description": "final velocity
"}}, "metadata": {"description": "Example 9 M1 book
", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equations question 10", "extensions": [], "custom_part_types": [], "resources": [["question-resources/ball_image.png", "/srv/numbas/media/question-resources/ball_image.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["u", "a", "t", "s"], "tags": [], "advice": "First it is best to sketch an image of what is happening. The ball rolls through $A$ with constant deceleration and a positive velocity. As it is decelerating it will eventually stop (have velocity $0 \\mathrm{ms^{-1}}$) at some unknown point, say $B$. However the ball will continue to decelerate and therefore will change direction and return to the point $A$.
\n![](\"resources/question-resources/ball_image.png\"/)
a) We have $u=\\var{u}$ and $a = \\var{a}$, where $a$ is negative as the ball is decelerating. We are asked to find the time, $t$, that the ball returns to point $A$, which means the balls displacement from $A$ will be zero. Therefore $s=0$. We use the formula $s = ut+\\frac{1}{2}at^2$ and solve for $t$. As this is a quadratic in $t$ there will be two solutions, however we know that the ball is at point $A$ when $t=0$ so we are trying to find the other value of $t$ which is when the ball returns to point $A$.
\nTherefore \\begin{align} s &= ut+ \\frac{1}{2}at^2, \\\\
\\frac{1}{2}at^2 + ut &= 0, \\\\
t\\left(\\frac{1}{2}at+u\\right) &= 0.\\end{align}
We discard the solution $t=0$ so \\begin{align}
\\frac{1}{2}at + u &=0, \\\\
t & = -\\frac{2u}{a}, \\\\
& = \\var{precround(-(2*u)/a,3)} \\mathrm{s}.
\\end{align}
Therefore the ball returns to point $A$ after $\\var{precround(-(2*u)/a,3)}$ seconds.
b) We have $u=\\var{u}$ and $a=\\var{a}$. At the point labelled $B$ in the diagram the velocity of the ball is $v=0$ for an instant as it has reached its furthest point from $A$ and will change direction. We want to find the distance between point $A$ and $B$ and then double it, to calculate the total distance the ball travels as it goes from $A$ to $B$ and back. Therefore we use the equation $v^2 = u^2+2as$, rearranged for $s$.
\\begin{align} v^2 & = u^2 + 2as, \\\\
2as & = v^2 - u^2, \\\\
s & = \\frac{v^2 - u^2}{2a}, \\\\
& = \\frac{ 0 - \\var{u}^2}{2 \\times \\var{a}}, \\\\
& = \\var{precround((-u^2)/(2*a),3)} \\mathrm{m}. \\end{align}
So we have the distance from point $A$ to point $B$, $s=\\var{precround((-u^2)/(2*a),3)}$ metres. Therefore the total distance travelled by the ball in $\\var{precround(-(2*u)/a,3)}$ seconds is $2 \\times s=\\var{precround(2*((-u^2)/(2*a)),3)}$ metres.
Find the time in $\\mathrm{s}$ (to 3d.p.) between when the ball first passes through point $A$ and when it returns to point $A$.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "-2u/a", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "-2u/a", "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "The total distance in $\\mathrm{m}$ travelled by the ball in this time.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "2*s", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "2*s", "type": "numberentry", "showPrecisionHint": false}], "statement": "At time $t=0$ a ball rolls through a point $A$ in the positive direction with a speed $\\var{u} \\mathrm{ms^{-1}}$. The ball rolls with constant deceleration $\\var{-a} \\mathrm{ms^{-2}}$.
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"css": "", "js": ""}, "variables": {"a": {"definition": "random(-5..-0.5#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "a", "description": "acceleration - negative as decelerating
"}, "s": {"definition": "-u^2/(2*a)", "templateType": "anything", "group": "Ungrouped variables", "name": "s", "description": ""}, "u": {"definition": "random(2..14#0.5)", "templateType": "randrange", "group": "Ungrouped variables", "name": "u", "description": "initial velocity of the ball
"}, "t": {"definition": "-2*u/a", "templateType": "anything", "group": "Ungrouped variables", "name": "t", "description": ""}}, "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equation question 11 - involves gravity", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["u"], "tags": [], "advice": "a) We will take the upwards direction as being the positive direction, therefore gravity will act in the negative direction with $a=-9.8 \\mathrm{ms^{-2}}$.
We have that $u=\\var{u}$ and at the greatest height above the ground the velocity of the rocket will be $v=0$ before it changes direction and heads back towards the ground. We want $s$ therefore we will use the formula $v^2=u^2+2as$, rearranged for $s$.
\\begin{align} v^2 &= u^2 + 2as, \\\\
2as &= v^2 - u^2, \\\\
s & = \\frac{v^2 - u^2}{2a}, \\\\
& = \\frac{0 - \\var{u}^2}{2\\times -9.8}, \\\\
& = \\var{precround(u^2/19.6,3)} \\mathrm{m}. \\end{align}
The greatest height reached by the rocket is $\\var{precround(u^2/19.6,3)}$ metres.
\nb) When the rocket returns to the ground, its displacement from the ground will be zero therefore $s=0$. We have $u=\\var{u}$ and $a=-9.8$ so we will use the formula $s=ut+\\frac{1}{2}at^2$ with $s=0$ to find the time it takes for the rocket to return to the ground.
\nTherefore \\begin{align} s & = ut+\\frac{1}{2}at^2, \\\\
& = t\\left(u + \\frac{1}{2}at\\right), \\end{align}
setting $s=0$ and discarding the solution $t=0$ as this represents the starting position of the rocket we obtain
\n\\begin{align} u + \\frac{1}{2}at &= 0, \\\\
t &=-\\frac{2u}{a}, \\\\
&= \\frac{2\\times \\var{u}}{9.8}, \\\\
&= \\var{precround((10/49)*u,3)} \\mathrm{s} . \\end{align}
The rocket returns to the ground after $\\var{precround((10/49)*u,3)}$ seconds.
\n", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "Find the greatest heightin $\\mathrm{m}$ (to 3d.p.) above the ground that is reached by the rocket.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "u^2/19.6", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "u^2/19.6", "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "The total time in $\\mathrm{s}$ (to 3d.p.) before the rocket returns to the ground.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "(10/49)*u", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "(10/49)*u", "type": "numberentry", "showPrecisionHint": false}], "statement": "A rocket is projected vertically upward from the ground with a speed $\\var{u} \\mathrm{ms^{-1}}$. Suppose that the acceleration due to gravity is $g= 9.8 \\mathrm{ms^{-2}}$.
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"css": "", "js": ""}, "variables": {"u": {"definition": "random(15..60#0.5)", "templateType": "randrange", "group": "Ungrouped variables", "name": "u", "description": ""}}, "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equations question 12", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["s", "a"], "tags": [], "advice": "We can model the plant pot as a particle moving in a straight line with a constant acceleration of $9.8ms^{-2}$, as it is accelerating under gravity. We will take the downward direction as the positive direction as the pot is falling downwards.
\na) We have $s=\\var{s}, a=\\var{a}$ and $u=0$ as we assume that the initial speed of the pot is zero. We want to find the time it hits the ground, $t$, therefore we can use the formula $s=ut+\\frac{1}{2}at^2$, with $u=0$, rearranged for $t$.
\nTherefore \\begin{align} s &= ut+\\frac{1}{2}at^2, \\\\
&= \\frac{1}{2}at^2, \\\\
t^2 & = \\frac{2s}{a}, \\\\
t & = \\sqrt{\\frac{2s}{a}}, \\\\
& = \\sqrt{\\frac{2\\times\\var{s}}{\\var{a}}},\\\\
& = \\var{precround(((2*s)/a)^(1/2),3)} \\mathrm{s}. \\end{align}
The plant pot hits the ground after $\\var{precround(((2*s)/a)^(1/2),3)}$ seconds.
b) We have $s=\\var{s}, u = 0$ and $a = 9.8$ and we want the speed with which the pot hits the ground, $v$. Therefore we can use the formula $v^2 = u^2 + 2as$.
\n\\begin{align} v^2 & = u^2 + 2as, \\\\
& = 0 + \\left(2 \\times \\var{a} \\times \\var{s}\\right), \\\\
& = \\var{precround(2*a*s,3)} \\mathrm{ms^{-1}} \\end{align}
Therefore $v = \\sqrt{\\var{precround(2*a*s,3)}} = \\var{precround((2*a*s)^(1/2),3)} \\mathrm{ms^{-1}}$ so the pot hits the ground at $\\var{precround((2*a*s)^(1/2),3)} \\mathrm{ms^{-1}}$.
", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "Find the time in $\\mathrm{s}$ (to 3d.p.) it takes for the pot to hit the ground.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "(2*s/a)^(1/2)", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "(2*s/a)^(1/2)", "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "Find the speed in $\\mathrm{ms^{-1}}$ (to 3d.p.) with which the pot hits the ground.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "(2*a*s)^(1/2)", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "(2*a*s)^(1/2)", "type": "numberentry", "showPrecisionHint": false}], "statement": "A plant pot falls off a window ledge which is $\\var{s} \\mathrm{m}$ above the garden floor. The acceleration due to gravity is $g= 9.8 \\mathrm{ms^{-2}}$
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"css": "", "js": ""}, "variables": {"a": {"definition": "9.8", "templateType": "number", "group": "Ungrouped variables", "name": "a", "description": "gravity acting downwards
"}, "s": {"definition": "random(0.5..4#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "s", "description": ""}}, "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equations question 13", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["a", "s", "u", "t", "t2", "max_height"], "tags": [], "preamble": {"css": "", "js": ""}, "advice": "You can model the ball as a particle moving in a straight line with constant acceleration of magnitude $9.8 \\mathrm{ms}^{-2}$ due to gravity. As the ball is thrown upwards, we take the positive direction as pointing upwards.
\nWe are told $u=\\var{u}$ and $a=\\var{a}$ and we are asked to find the greatest height, $s$, reached by the ball. At its greatest height, the ball will for an instant have a velocity of $v=0$ before it starts to fall back down.
\nTherefore we can use the equation $v^2=u^2+2as$, rearranged for $s$.
\nNote that the rooftop is $\\var{s} \\mathrm{m}$ above ground level, so the distance travelled by the ball to its greatest height is $(\\simplify{s-{s}}) \\mathrm{m}$.
\n\\begin{align} v^2 & =u^2+2a(s-\\var{s}), \\\\
2a(s-\\var{s}) & = v^2-u^2, \\\\
s-\\var{s} & = \\frac{v^2-u^2}{2a}, \\\\
& = \\frac{0 - \\var{u}^2}{2 \\times \\var{a}}, \\\\
& = \\var{precround(-u^2/(2*a),3)}, \\\\
s &= \\simplify[]{{s}+{precround(-u^2/(2*a),3)}}, \\\\
&= \\var{precround(max_height,3)} \\mathrm{m}.
\\end{align}
The greatest height reached by the ball is $\\var{precround(max_height,3)} \\mathrm{m}$.
\nThe flight time of the ball is the length of time between it being thrown from the rooftop and when it hits the ground below.
\nHere, the rooftop is $\\var{s}\\mathrm{m}$ above the ground, therefore the ball will stop moving when it hits the ground at a height $\\var{-s} \\mathrm{m}$ from its original position.
\nTherefore we can use the equation $s=ut+\\frac{1}{2}at^2$ rearranged for $t$.
\n\\begin{align} s & = ut + \\frac{1}{2}at^2, \\\\
-\\var{s} & = \\var{u}t - \\frac{49}{10}t^2, \\\\
\\frac{49}{10}t^2 - \\var{u}t - \\var{s} &= 0.
\\end{align}
Using the quadratic formula, taking only the positive term as we are modelling time, this gives
\n\\begin{align} t & = \\frac{u + \\sqrt{u^2 +19.6s)}}{9.8}, \\\\
& = \\frac{\\var{u} + \\sqrt{\\var{u^2} + \\left(19.6 \\times \\var{s}\\right)}}{9.8}, \\\\
& = \\var{precround((u+ (u^2 + 19.6*s)^(1/2))/9.8,3)} \\mathrm{s}.
\\end{align}
The flight time of the ball is $\\var{precround((u+ (u^2 + 19.6*s)^(1/2))/9.8,3)}$ seconds.
", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "Find the greatest height in $\\mathrm{m}$ (to 3d.p.) above the ground reached by the ball.
", "precisionMessage": "You have not given your answer to the correct precision.
", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "max_height", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "max_height", "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "The time in $mathrm{s}$ (to 3d.p.) of flight of the ball, before it lands on the ground.
", "precisionMessage": "You have not given your answer to the correct precision.
", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "t2", "variableReplacementStrategy": "originalfirst", "strictPrecision": true, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "minValue": "t2", "type": "numberentry", "showPrecisionHint": false}], "statement": "A ball is thrown vertically upwards with a speed $\\var{u} \\mathrm{ms^{-1}}$ from a rooftop which is $\\var{s} \\mathrm{m}$ above ground. The acceleration due to gravity is $g = 9.8 \\mathrm{ms^{-2}}$.
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"a": {"definition": "-9.8", "templateType": "number", "group": "Ungrouped variables", "name": "a", "description": "Acceleration due to gravity with upwards as positive direction
"}, "t2": {"definition": "precround(t,3)", "templateType": "anything", "group": "Ungrouped variables", "name": "t2", "description": "time to 3d.p.
"}, "max_height": {"definition": "u^2/(2*(-a))+s", "templateType": "anything", "group": "Ungrouped variables", "name": "max_height", "description": ""}, "s": {"definition": "random(5..15#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "s", "description": "distance of rooftop above ground
"}, "u": {"definition": "random(10..30#0.5)", "templateType": "randrange", "group": "Ungrouped variables", "name": "u", "description": "initial velocity ball is thrown at
"}, "t": {"definition": "(u+(u^2+19.6*s)^(1/2))/9.8", "templateType": "anything", "group": "Ungrouped variables", "name": "t", "description": ""}}, "metadata": {"description": "A ball is thrown vertically upwards from a position above ground level. Find its greatest height, and the total time its in the air.
", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SUVAT equations question 14", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["s", "a", "s2", "distance", "u", "t1", "t2"], "tags": [], "advice": "We are told that the ball is thrown upwards, so let the upwards direction be positive and gravity will act in the opposite direction, thus $a=-9.8$. At the ball's greatest height, $s=\\var{s}$, the velocity will be $v=0$ for an instant as the ball changes direction and falls downwards. We are trying to find the initial velocity of the ball, $u$. We have everything except $t$ so we can use the formula $v^2 = u^2 + 2as$, rearranged for $u$.
\n\\begin{align} v^2 & = u^2 + 2as, \\\\
0 & = u^2 + \\left(2 \\times \\var{a} \\times \\var{s}\\right), \\\\
0 & = u^2 - \\var{2*-a*s}.
\\end{align}
Therefore
\n\\begin{align} u^2 &= \\var{2*-a*s}, \\\\
u & = \\var{precround ((2*-a*s)^(1/2),3)} \\mathrm{ms^{-1}}.
\\end{align}
The initial velocity of the ball is $ \\var{precround ((2*-a*s)^(1/2),3)} \\mathrm{ms^{-1}}.$
\nWe have now that $s=\\var{distance}$, $u = \\var{u}$ and $a=\\var{a}$. To find the time the ball is above $\\var{distance} \\mathrm{m}$ we need to find the time the ball first passes over $\\var{distance} \\mathrm{m}$ and the time it then falls back down past $\\var{distance} \\mathrm{m}$, and find the difference between those two times. To find these times we can use the formula $s=ut+\\frac{1}{2}at^2$, rearranged for $t$.
\nWe have
\n\\begin{align} s & = ut + \\frac{1}{2}at^2, \\\\
\\var{distance} & = \\var{u}t - \\left( \\frac{1}{2} \\times \\var{-a} \\times t^2 \\right), \\\\
4.9t^2 - \\var{u}t + \\var{distance} & = 0.
\\end{align}
Therefore our first $t$ value, when the ball first passes $\\var{distance} \\mathrm{m}$, is
\n\\begin{align} t & = \\frac{\\var{u} + \\sqrt{\\var{u}^2 - \\left(4 \\times 4.9 \\times \\var{distance} \\right)} }{2 \\times 4.9}, \\\\
& = \\var{t1} \\mathrm{s},
\\end{align}
and our second $t$ value, when the ball second passes $\\var{distance} \\mathrm{m}$, is
\n\\begin{align} t & = \\frac{\\var{u} - \\sqrt{\\var{u}^2 - \\left(4 \\times 4.9 \\times \\var{distance} \\right)} }{2 \\times 4.9}, \\\\
& = \\var{t2} \\mathrm{s}.
\\end{align}
Then the total time for which the ball is $\\var{distance} \\mathrm{m}$ or more above $X$ is $\\var{t1} - \\var{t2} = \\var{t1-t2}$ seconds.
", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "Find the initial speed in $\\mathrm{ms^{-1}}$ (to 3 decimal places) of the ball when it is thrown.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "(2*9.8*s)^(1/2)", "minValue": "(2*9.8*s)^(1/2)", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "Find the total time in $\\mathrm{s}$ for which the ball is $\\var{s-(15+s2)} \\mathrm{m}$ or more above $X$.
\nGive your answer to 3 decimal places.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "t1-t2+0.02", "minValue": "t1-t2-0.02", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "A ball is thrown vertically upwards from a point $X$. The greatest height reached by the ball is $\\var{s} \\mathrm{m}$ above $X$. The acceleration due to gravity is $g = 9.8 \\mathrm{ms^{-2}}$.
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": "distance>0"}, "preamble": {"css": "", "js": ""}, "variables": {"a": {"definition": "-9.8", "templateType": "number", "group": "Ungrouped variables", "name": "a", "description": ""}, "distance": {"definition": "s-(15+s2)", "templateType": "anything", "group": "Ungrouped variables", "name": "distance", "description": "total time for which the ball is distance or more above X
"}, "s2": {"definition": "random(0..10#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "s2", "description": ""}, "t2": {"definition": "precround((u-(u^2-19.6*distance)^(1/2))/9.8,3)", "templateType": "anything", "group": "Ungrouped variables", "name": "t2", "description": ""}, "t1": {"definition": "precround((u+(u^2-19.6*distance)^(1/2))/9.8,3)", "templateType": "anything", "group": "Ungrouped variables", "name": "t1", "description": ""}, "s": {"definition": "random(40..80#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "s", "description": "greatest height reached by the ball
"}, "u": {"definition": "precround((2*9.8*s)^(1/2),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "u", "description": ""}}, "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}]}], "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "extensions": [], "custom_part_types": [], "resources": [["question-resources/Car_image.jpg", "/srv/numbas/media/question-resources/Car_image.jpg"], ["question-resources/truck_image.png", "/srv/numbas/media/question-resources/truck_image.png"], ["question-resources/man_image.png", "/srv/numbas/media/question-resources/man_image.png"], ["question-resources/ball_image.png", "/srv/numbas/media/question-resources/ball_image.png"]]}