// Numbas version: finer_feedback_settings {"name": "SUVAT equations question 1", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Car_image.jpg", "/srv/numbas/media/question-resources/Car_image.jpg"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "u", "t", "v"], "name": "SUVAT equations question 1", "tags": [], "advice": "

You start by writing down the values you know and the values you need to find.

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$a = \\var{a},$ $u=\\var{u},$ $t=\\var{t},$ $v=$ $?,$ $s=$ $?$

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a) 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}

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The speed of the car at $B$ is $\\var{u + a*t} \\mathrm{ms^{-1}}.$

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b) You need the distance, $s \\mathrm{m}$. You calculated $v$ in the previous part, so you can use the equation $s=\\left(\\frac{u+v}{2}\\right)t.$

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\\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.

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Find the speed of the car in $\\mathrm{ms^{-1}}$ at $B$.

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Find the distance in $\\mathrm{m}$ from $A$ to $B$.

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A car is driving in a straight line from $A$ to $B$ with constant acceleration $\\var{a} \\mathrm{ms^{-2}}$.

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Its speed at $A$ is $\\var{u} \\mathrm{ms^{-1}}$ and it takes $\\var{t}$ seconds to move from $A$ to $B$.

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constant acceleration

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initial speed

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time (seconds)

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part a) answer. final velocity

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SUVAT equation questions, for mechanics page in wiki. Uses $v=u+at$ and $s=((u+v)/2)t$.

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