A ball rolls in a straight horizontal line with constant deceleration $\\var{-a} \\mathrm{ms^{-2}}.$ 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 $A$.
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", "allowFractions": false, "minValue": "t2", "showFeedbackIcon": true, "mustBeReduced": false, "precisionPartialCredit": 0, "strictPrecision": false, "unitTests": [], "notationStyles": ["plain", "en", "si-en"], "showPrecisionHint": false, "marks": 1, "extendBaseMarkingAlgorithm": true, "correctAnswerFraction": false, "variableReplacements": [], "precision": "3", "correctAnswerStyle": "plain", "type": "numberentry", "maxValue": "t2", "precisionType": "dp", "showCorrectAnswer": true, "precisionMessage": "You have not given your answer to the correct precision.", "scripts": {}, "mustBeReducedPC": 0, "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst"}, {"prompt": "Find the longest time in $\\mathrm{s}$ (to 3d.p.) when the ball rolls through $B$. This will be in the direction $\\simplify{vec:BA}$.
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", "allowFractions": false, "minValue": "(-2*u)/a", "showFeedbackIcon": true, "mustBeReduced": false, "precisionPartialCredit": 0, "strictPrecision": false, "unitTests": [], "notationStyles": ["plain", "en", "si-en"], "showPrecisionHint": false, "marks": 1, "extendBaseMarkingAlgorithm": true, "correctAnswerFraction": false, "variableReplacements": [], "precision": "3", "correctAnswerStyle": "plain", "type": "numberentry", "maxValue": "(-2*u)/a", "precisionType": "dp", "showCorrectAnswer": true, "precisionMessage": "You have not given your answer to the correct precision.", "scripts": {}, "mustBeReducedPC": 0, "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst"}], "preamble": {"css": "", "js": ""}, "name": "Simon's copy of SUVAT equations question 8", "tags": [], "rulesets": {}, "advice": "a) & b) We have $a=\\var{a}$, $u=\\var{u}$ and $s=\\var{s}$. The ball is decelerating so acceleration is negative.
\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 $t=\\var{precround(t2,4)}$ we get
\\begin{align} v &= u + at, \\\\
&=\\var{u} + \\left(\\var{a}\\times\\var{precround(t2,3)}\\right), \\\\
&= \\var{precround(u+a*precround(t2,4),3)} \\mathrm{ms^{-1}}. \\end{align}
This is positive, therefore the ball is travelling at $\\var{precround(u+a*precround(t2,4),3)} \\mathrm{ms^{-1}}$ in the direction $\\simplify{vec:AB}$.
\nd) Following the same method as part c) only now using $t=\\var{precround(t1,4)}$ we get
\\begin{align} v &= u + at, \\\\
&=\\var{u} + \\left(\\var{a}\\times\\var{precround(t1,4)}\\right), \\\\
&= \\var{precround(u+a*precround(t1,4),3)} \\mathrm{ms^{-1}}. \\end{align}
This is negative, therefore the ball is travelling at $\\var{-(precround(u+a*precround(t1,4),3))} \\mathrm{ms^{-1}}$ 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 $\\var{precround(-(2/a)*u,3)} \\mathrm{s}$.
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