// Numbas version: exam_results_page_options {"name": "Constant Acceleration: Rearranging Equations Quiz", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], []], "questions": [{"name": "Constant Acceleration: Rearranging Equations 1a", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}], "tags": [], "metadata": {"description": "

Rearraning the constant acceleration equation $v=u+at$ to make $a$ the subject.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rearrange the constant acceleration equation \\[ v = u+ at \\] to make $a$ the subject.

", "advice": "

If we want to make $a$ the subject of the equation $v=u+at$, this can be broken down into two steps.

Firstly, we want to subtract $u$ from both sides:

\\[ \\begin{split} &\\,v=u+at \\\\\\\\ \\implies &\\,v-u =at \\,.\\end{split} \\]

Now the right-hand side of the equation is only the product of $a$ and $t$, so to get $a$ on its own, we want to divide both sides of the equation by $t$ :

\\[ \\begin{split} v-u &\\,=at \\\\\\\\ \\implies \\frac{v-u}{t} &\\,=a \\,.\\end{split} \\]

Therefore, the equation with $a$ as the subject is \\[ a=\\frac{v-u}{t} \\,.\\]

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$a=$[[0]]

(Remember to use a * when multiplying terms together. For example, ab = a*b)

Rearraning the constant acceleration equation $v=u+at$ to make $t$ the subject.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rearrange the constant acceleration equation \\[ v = u+ at \\] to make $t$ the subject.

", "advice": "

If we want to make $t$ the subject of the equation $v=u+at$, this can be broken down into two steps.

Firstly, we want to subtract $u$ from both sides:

\\[ \\begin{split} &\\,v=u+at \\\\\\\\ \\implies &\\,v-u =at \\,.\\end{split} \\]

Now the right-hand side of the equation is only the product of $a$ and $t$, so to get $t$ on its own, we want to divide both sides of the equation by $a$ :

\\[ \\begin{split} v-u &\\,=at \\\\\\\\ \\implies \\frac{v-u}{a} &\\,=t \\,.\\end{split} \\]

Therefore, the equation with $t$ as the subject is \\[ t=\\frac{v-u}{a} \\,.\\]

$t=$[[0]]

(Remember to use a * when multiplying terms together. For example, ab = a*b)

Rearraning the constant acceleration equation $s=ut+\\frac{1}{2}at^2$ to make $u$ the subject.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rearrange the constant acceleration equation \\[ s = ut+ \\frac{1}{2}at^2 \\] to make $u$ the subject.

", "advice": "

If we want to make $u$ the subject of the equation $s=ut+\\frac{1}{2}at^2$, this can be broken down into two steps.

Firstly, we want to subtract $\\frac{1}{2}at^2$ from both sides:

\\[ \\begin{split} &\\,s=ut+\\frac{1}{2}at^2 \\\\\\\\ \\implies &\\,s-\\frac{1}{2}at^2 =ut \\,.\\end{split} \\]

Now the right-hand side of the equation is only the product of $u$ and $t$, so to get $u$ on its own, we want to divide both sides of the equation by $t$ :

\\[ \\begin{split} s-\\frac{1}{2}at^2 &\\,=ut \\\\\\\\ \\implies \\frac{s-\\frac{1}{2}at^2}{t} &\\,=u \\,.\\end{split} \\]

Therefore, the equation with $u$ as the subject is \\[ u=\\frac{s-\\frac{1}{2}at^2}{t} \\,,\\]

which can be rewritten as \\[ u = \\frac{s}{t} - \\frac{1}{2}at \\,.\\]

$u=$[[0]]

(Remember to use a * when multiplying terms together. For example, ab = a*b)

Rearraning the constant acceleration equation $s=ut+\\frac{1}{2}at^2$ to make $a$ the subject.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rearrange the constant acceleration equation \\[ s = ut+ \\frac{1}{2}at^2 \\] to make $a$ the subject.

", "advice": "

If we want to make $a$ the subject of the equation $s=ut+\\frac{1}{2}at^2$, this can be broken down into steps.

Firstly, we want to subtract $ut$ from both sides:

\\[ \\begin{split} &\\,s=ut+\\frac{1}{2}at^2 \\\\\\\\ \\implies &\\,s-ut=\\frac{1}{2}at^2 \\,.\\end{split} \\]

Now the right-hand side of the equation is the product of $\\frac{1}{2}$, $a$ and $t^2$, so we can now divide both sides of the equation by $t^2$:

\\[ \\begin{split} s-ut &\\,=\\frac{1}{2}at^2 \\\\\\\\ \\implies \\frac{s-ut}{t^2} &\\,=\\frac{1}{2}a \\,.\\end{split} \\]

Since we now have an expression for $\\frac{1}{2}a$, to get $a$ on its own we want to multiply both sides of the equation by 2:

\\[ \\begin{split} \\frac{s-ut}{t^2} &\\,=\\frac{1}{2}a \\\\\\\\ 2\\frac{(s-ut)}{t^2} &\\,=a \\,. \\end{split} \\]

Therefore, the equation with $a$ as the subject is \\[ a=2\\frac{(s-ut)}{t^2} \\,.\\]

Note: This is $\\mathbf {not}$ the only way of writing this equation in terms of $a$:

\\[ a = 2\\frac{(s-ut)}{t^2} \\equiv 2\\left(\\frac{s}{t^2}-\\frac{u}{t} \\right) \\equiv \\frac{2s}{t^2}-\\frac{2u}{t} \\,. \\]

$\\mathbf {All\\,\\, of\\,\\, these\\,\\, answers\\,\\, are\\,\\, correct!}$

$a=$[[0]]

(Remember to use a * when multiplying terms together. For example, ab = a*b)

Rearraning the constant acceleration equation $v^2=u^2+2as$ to make $u$ the subject.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rearrange the constant acceleration equation \\[ v^2=u^2+2as \\] to make $u$ the subject.

", "advice": "

If we want to make $u$ the subject of the equation $v^2=u^2+2as$, this can be broken down into steps.

Firstly, we want to subtract $2as$ from both sides:

\\[ \\begin{split} &\\,v^2=u^2+2as \\\\\\\\ \\implies &\\,v^2-2as=u^2 \\,.\\end{split} \\]

Now the right-hand side of the equation is only $u^2$, so by taking the square root of both sides we can obtain an expression for $u$:

\\[ \\begin{split}v^2-2as=u^2 \\\\\\\\ \\implies \\sqrt{v^2-2as} &\\,=u \\,.\\end{split} \\]

Therefore, the equation with $u$ as the subject is \\[ u=\\sqrt{v^2-2as} \\,.\\]

$u=$[[0]]

If you need to use a square root, type \"sqrt\". For example, $\\sqrt{x+y}=$ \"sqrt(x+y)\"
Remember to use a * when multiplying terms together. For example, ab = a*b

Rearraning the constant acceleration equation $v^2=u^2+2as$ to make $a$ the subject.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rearrange the constant acceleration equation \\[ v^2=u^2+2as \\] to make $a$ the subject.

", "advice": "

If we want to make $a$ the subject of the equation $v^2=u^2+2as$, this can be broken down into steps.

Firstly, we want to subtract $u^2$ from both sides:

\\[ \\begin{split} &\\,v^2=u^2+2as \\\\\\\\ \\implies &\\,v^2-u^2=2as \\,.\\end{split} \\]

Now the right-hand side is the product of $2$, $a$ and $s$, so to get an expression for $a$ we want to divide both sides of the equation by $2s$:

\\[ \\begin{split}v^2-u^2=2as \\\\\\\\ \\implies \\frac{v^2-u^2}{2s} &\\,=a \\,.\\end{split} \\]

Therefore, the equation with $a$ as the subject is \\[ a=\\frac{v^2-u^2}{2s} \\,.\\]

$a=$[[0]]

(Remember to use a * when multiplying terms together. For example, ab = a*b)

Rearraning the constant acceleration equation $v^2=u^2+2as$ to make $s$ the subject.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rearrange the constant acceleration equation \\[ v^2=u^2+2as \\] to make $s$ the subject.

", "advice": "

If we want to make $s$ the subject of the equation $v^2=u^2+2as$, this can be broken down into steps.

Firstly, we want to subtract $u^2$ from both sides:

\\[ \\begin{split} &\\,v^2=u^2+2as \\\\\\\\ \\implies &\\,v^2-u^2=2as \\,.\\end{split} \\]

Now the right-hand side is the product of 2, $a$ and $s$, so to get an expression for $s$ we want to divide both sides of the equation by $2a$:

\\[ \\begin{split}v^2-u^2=2as \\\\\\\\ \\implies \\frac{v^2-u^2}{2a} &\\,=s \\,.\\end{split} \\]

Therefore, the equation with $s$ as the subject is \\[ s=\\frac{v^2-u^2}{2a} \\,.\\]

$s=$[[0]]

(Remember to use a * when multiplying terms together. For example, ab = a*b)

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