// Numbas version: finer_feedback_settings {"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": [{"name": "Using a speed graph to find distance", "tags": [], "metadata": {"description": "

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

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

", "licence": "Creative Commons Attribution 4.0 International"}, "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 (m/s or ms$^{-1}$).

{geogebra_applet('cecdYjwp',[[\"a1\",a1],[\"b1\",b1],[\"c1\",c1],[\"d1\",d1],[\"z1\",z1],[\"f1\",f1]])}

", "advice": "

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.

a)

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.

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

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

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

b)

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

\\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}

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

c)

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 $B$).

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

\\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}

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

\\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}

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

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

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

d)

The greatest acceleration in a velocity-time graph will be the part of the graph with the highest (positive) slope. Our graph is 'climbing' upwards between A and B and again between B and C.

Slope between A and B:

\\begin{align}
m_{ab} &= \\frac{\\var{b1}-\\var{a1}}{2-0}\\\\
&= \\var{mab}\\\\
\\end{align}

Slope between B and C:

\\begin{align}
m_{ab} &= \\frac{\\var{c1}-\\var{b1}}{4-2}\\\\
&= \\var{mbc}\\\\
\\end{align}

So the greatest acceleration is the higher of the two: {maxm} m/s²

e)

Speed is the distance travelled per unit of time.

\\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}

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

Distance travelled $=$ [[0]] m

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

Distance travelled $=$ [[0]] m

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

Distance travelled $=$ [[0]] m

What was the value of the greatest acceleration of the car?

Acceleration = [[0]] m/s²

The car travelled {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.

Average speed $=$ [[0]] m/s

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