// Numbas version: exam_results_page_options {"name": "Using a speed graph to find distance", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {}, "variable_groups": [{"variables": ["a1", "b1", "c1", "d1", "z1", "f1"], "name": "speeds"}, {"variables": ["mab", "mbc", "mcd", "mde", "mef"], "name": "acceleration"}], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Use a piecewise linear graph of speed against time to find the distance travelled by a car.

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Finally, use the total distance travelled to find the average speed.

"}, "tags": ["average speed", "calculating distance", "constant accelerations", "distance", "Distance", "graphs", "speed graph", "taxonomy"], "functions": {}, "name": "Using a speed graph to find distance", "ungrouped_variables": ["d2farea", "area"], "parts": [{"variableReplacementStrategy": "originalfirst", "scripts": {}, "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "gapfill", "prompt": "

Use the graph to calculate the distance the car travels between $4$ and $6$ seconds.

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Distance travelled $=$  [[0]]metres

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Use the graph to calculate the distance the car travels between $0$ and $2$ seconds.

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Distance travelled $=$  [[0]]metres

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Use the graph to calculate the distance the car travels between $2$ and $4$ seconds.

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Distance travelled $=$  [[0]]metres

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The car travelled $\\var{area}$ metres over the $10$ second period. Calculate the average speed of the the car over the $10$ seconds in metres per second. Give your answer as a whole number or a decimal to $1$ decimal place.

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Average speed $=$ [[0]] ms-1

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We can use a speed graph to calculate the distance travelled in a given time interval by finding the area under the line between the start and end times.

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a)

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The shape made by the speed curve, the line $x=0$, and the lines $t=4$ and $t=6$ seconds is a rectangle, so we can work out the area of this section by multiplying the width by the height.

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The rectangle is $2$ seconds wide, and $\\var{c1}$ ms-1 high.

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\\begin{align}
\\text{Area} &= \\text{width} \\times \\text{height}\\\\
&= 2 \\times\\var{c1}\\\\
&=\\simplify{2{c1}}\\text{.}
\\end{align}

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So the distance covered in this two second interval is $\\simplify{2{c1}}$ m.

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b)

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The shape made by the line and $x=0$ between $0$ and $2$ seconds forms a right-angled triangle with width $2$ and height $\\var{b1}$.

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\\begin{align}
\\text{Area}&=  \\frac{1}{2}\\times \\text{width} \\times \\text{height}\\\\
&= \\frac{1}{2} \\times 2 \\times \\var{b1}\\\\
&=\\var{b1} \\text{.}
\\end{align}

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So therefore,the distance covered in this two second interval, and our answer, is $\\simplify{{b1}}$ meters.

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c)

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The shape made by the speed curve and $x=0$ between $2$ and $4$ seconds forms a trapezium. This can be broken down in to a right angle triangle (let's call this $A$) and a rectangle (we'll call this $B$).

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Triangle $A$ has width $2$ m and height $\\var{c1}-\\var{b1}$ ms-1.

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\\begin{align}
A &=  \\frac{1}{2}\\times \\text{width} \\times \\text{height}\\\\
&= \\frac{1}{2}\\times2 \\times\\ (\\var{c1}-\\var{b1})\\\\
&= \\var{c1}-\\var{b1}\\\\
&=\\simplify{{c1}-{b1}}\\text{.}
\\end{align}

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We can work out the area of the rectangle $B$ by multiplying its width, $2$ seconds, by its height, $\\var{b1}$ ms-1:

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\\begin{align}
B &=  \\text{width} \\times \\text{height}\\\\
&= 2 \\times(\\var{c1}-\\var{b1})\\\\
&=2 \\times \\simplify{{c1}-{b1}}\\\\
&=\\simplify{2{c1-b1}}\\text{.}
\\end{align}

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We can now work out the whole area under the line by adding these two areas together:

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\\begin{align}
\\text{Area} &= A + B \\\\
&=\\simplify{{c1}-{b1}} + \\simplify{2{c1-b1}} \\\\
&=\\simplify{2{c1-b1}+{c1}-{b1}} \\text{.}
\\end{align}

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The distance covered in this interval is $\\var{2(c1-b1)+c1-b1}$ m.

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d)

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Speed is the distance travelled per unit of time.

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\\begin{align}
\\text{speed} &= \\frac{\\text{distance}}{\\text{time}} \\\\[0.5em]
&= \\frac{\\var{area}}{10} \\\\[0.5em]
&=\\simplify[!fractionNumbers]{{area/10}} \\text{ ms}^{-1}\\text{.}
\\end{align}

", "statement": "

You are part of an elite team analysing a high speed car race. You are given the following graph mapping the speed of one particular car as it drives around a section of the race course. The horizontal axis plots time in seconds whilst the vertical axis maps speed in metres per second ($ms^{-1}$).

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