Question asks student to find zeros of a quadratic equation - disguised as finding time for particle to reach a given position. In this version students are expected to write up their working and submit it seperately to the Numbas question. Students are expcted to recognise that only the positive solution has physical significance. Coefficients of the quadratic are randomly chosen within linits which give one positive and one negative root.
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\nUsing the quadratic formula we have \\begin{align} t&=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\\\&=\\frac{-\\var{a}\\pm\\sqrt{\\simplify[!collectNumbers]{{a}^2-4*{b}*{-c}}}}{\\var{2*b}}\\\\&=\\frac{-\\var{a}+\\sqrt{\\simplify{{a}^2-4*{b}*{-c}}}}{\\var{2*b}}\\text{ or }\\frac{-\\var{a}-\\sqrt{\\simplify{{a}^2-4*{b}*{-c}}}}{\\var{2*b}}\\\\&=\\var{ans}\\text{ or }\\var{ans2}\\end{align}
\nSince we are told that \\(t\\ge 0\\) we can ignore the negative answer and conclude that particle will have a displacement of \\(\\var{c}\\) m when \\(t\\) is approximately \\(\\var{precround({ans},2)}\\) seconds.
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\nFind the time taken for the particle to have a displacement of \\(\\var{c}\\) m. Write your final answer in the box below and show full working on your handwritten notes.
\n\\(t=\\)[[0]]sec (round answer to 2 decimal places)
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