// Numbas version: finer_feedback_settings
{"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}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["u", "a", "t", "s"], "name": "SUVAT equations question 10", "tags": [], "advice": "<p>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 $A$.</p>\n<p><img width=\"386\" height=\"166\" style=\"margin-right: auto; margin-left: auto; display: block;\" src=\"resources/question-resources/ball_image.png\"/></p>\n<p>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, $t$,&nbsp;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 <em>returns</em> to point $A$.</p>\n<p>Therefore \\begin{align} s &amp;= ut+ \\frac{1}{2}at^2, \\\\<br/>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \\frac{1}{2}at^2 + ut &amp;= 0, \\\\<br/>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; t\\left(\\frac{1}{2}at+u\\right) &amp;= 0.\\end{align}</p>\n<p>We discard the solution $t=0$ so&nbsp;\\begin{align}<br/>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;\\frac{1}{2}at + u&nbsp;&amp;=0, \\\\<br/>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; t&nbsp;&amp; = -\\frac{2u}{a}, \\\\<br/>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&amp; =&nbsp;\\var{precround(-(2*u)/a,3)} \\mathrm{s}.<br/>\\end{align}&nbsp;<br/>Therefore the ball returns to point $A$ after $\\var{precround(-(2*u)/a,3)}$ seconds.</p>\n<p>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 <em>double</em> it, to calculate the total distance the ball travels as it goes from $A$ to $B$ and <em>back</em>. Therefore we use the equation $v^2 = u^2+2as$, rearranged for $s$.<br/>\\begin{align} v^2 &amp; = u^2 + 2as, \\\\<br/>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2as &amp; = v^2 - u^2, \\\\<br/>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;s &amp; = \\frac{v^2 - u^2}{2a}, \\\\<br/>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &amp; = \\frac{ 0 - \\var{u}^2}{2 \\times \\var{a}}, \\\\<br/>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &amp; = \\var{precround((-u^2)/(2*a),3)} \\mathrm{m}. \\end{align}<br/>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.</p>", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "<p>Find the time in $\\mathrm{s}$ (to 3d.p.)&nbsp;between when the ball first passes through point $A$ and when it returns to point $A$.</p>", "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": "<p>The total distance in $\\mathrm{m}$ travelled by the ball in this time.</p>", "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}], "extensions": [], "statement": "<p>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 $\\var{-a} \\mathrm{ms^{-2}}$.</p>", "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": "<p>acceleration - negative as decelerating</p>"}, "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": "<p>initial velocity of the ball</p>"}, "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": []}], "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}]}]}], "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}]}