// Numbas version: exam_results_page_options {"question_groups": [{"name": "Group", "pickQuestions": 1, "pickingStrategy": "all-ordered", "questions": [{"name": "Using a speed and acceleration graph", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Bradley Bush", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1521/"}], "variable_groups": [{"variables": ["a1", "b1", "c1", "d1", "z1", "f1"], "name": "speeds"}, {"variables": ["mab", "mbc", "mcd", "mde", "mef"], "name": "acceleration"}], "preamble": {"js": "", "css": ""}, "type": "question", "parts": [{"variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "gaps": [{"maxMarks": 0, "type": "1_n_2", "showCorrectAnswer": true, "minMarks": 0, "distractors": ["", ""], "variableReplacementStrategy": "originalfirst", "displayType": "dropdownlist", "shuffleChoices": false, "showFeedbackIcon": true, "displayColumns": 0, "matrix": [0, "1"], "choices": ["

speed

", "

acceleration

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speed

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acceleration

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The green line with diamond shaped points maps the [[0]] of the car whilst the blue line with circular points maps the [[1]] of the car.

"}, {"variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "gaps": [{"maxMarks": 0, "type": "1_n_2", "showCorrectAnswer": true, "minMarks": 0, "distractors": ["", ""], "variableReplacementStrategy": "originalfirst", "displayType": "radiogroup", "shuffleChoices": false, "showFeedbackIcon": true, "displayColumns": 0, "matrix": [0, "1"], "choices": ["

Yes - a car could have a negative speed.

", "

No - a car can never have a negative speed.

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The graph shows no negative speeds; is there ever a case where the car could've had a negative speed? [[0]]

Yes - a car could have a negative acceleration.

", "

No - a car can never have a negative acceleration.

"], "scripts": {}, "variableReplacements": [], "marks": 0}], "variableReplacements": [], "showFeedbackIcon": true, "prompt": "

The graph shows some negative accelerations; is it possible for a car to have negative acceleration? [[0]]

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Calculate acceleration over the first $2$ seconds. Use the acceleration line on the graph to check your answer is correct.

Acceleration $=$ [[0]] $m\\space/\\space s^2$

"}], "advice": "

a)

We can analyse the nature of each line to find out which line resembles speed and which resembles acceleration.

The speed line should be distinguishable from the acceleration line based on the fact speed is always positive and has a continuous function (a line where the points are joined). This suggests the blue line with circular points represents speed.

Where as acceleration can often be discontinuous and can be negative where speeds are decreasing. This suggests that the green line with diamond shaped points is the acceleration line.

b)

Speed is a scalar quantity without direction so speed is positive no matter the direction it travels in. Therefore our answer is \"no- a car can never have a negative speed.\".

c)

By defintion, when an object is slowing down it has a negative acceleration as the acceleration measures that change in speed. Therefore, our answer is \"yes- a car could have negative acceleration.\".

d)

To calculate acceleration we use the point at $0$ seconds, with a $0$ m/s speed and the point at $2$ seconds with a $\\var{b1}$ m/s speed and we find the gradient of the line between them. To find the gradient of a straight line we can use:

\\[
\\begin{align}
\\text{Gradient } &= \\frac{y_1-y_2}{x_1-x_2}\\\\
&= \\frac{\\var{b1-a1}}{2-0}\\\\
&=\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{b1-a1}/2} \\text{.}
\\end{align}
\\]

We can then compare the graph to our this answer in order to check that our calculation is correct and we see that we the same answer as was given.

", "tags": ["acceleration", "graphs", "speed and acceleration graph", "taxonomy"], "variables": {"area": {"templateType": "anything", "description": "", "definition": "d2farea+c1*2+b1+c1+b1", "name": "area", "group": "Ungrouped variables"}, "d2farea": {"templateType": "anything", "description": "", "definition": "2*z1+d1+f1", "name": "d2farea", "group": "Ungrouped variables"}, "mab": {"templateType": "anything", "description": "", "definition": "(b1-a1)/2", "name": "mab", "group": "acceleration"}, "b1": {"templateType": "anything", "description": "", "definition": "random(6,8,10)+a1\n", "name": "b1", "group": "speeds"}, "mef": {"templateType": "anything", "description": "", "definition": "(f1-z1)/2", "name": "mef", "group": "acceleration"}, "a1": {"templateType": "anything", "description": "", "definition": "0\n", "name": "a1", "group": "speeds"}, "d1": {"templateType": "anything", "description": "", "definition": "c1", "name": "d1", "group": "speeds"}, "f1": {"templateType": "anything", "description": "", "definition": "2", "name": "f1", "group": "speeds"}, "c1": {"templateType": "anything", "description": "", "definition": "random(8,10,12,14)+b1", "name": "c1", "group": "speeds"}, "mde": {"templateType": "anything", "description": "", "definition": "(z1-d1)/2", "name": "mde", "group": "acceleration"}, "mbc": {"templateType": "anything", "description": "", "definition": "(c1-b1)/2", "name": "mbc", "group": "acceleration"}, "mcd": {"templateType": "anything", "description": "", "definition": "(d1-c1)/2", "name": "mcd", "group": "acceleration"}, "z1": {"templateType": "anything", "description": "", "definition": "c1-random(8,10,12)", "name": "z1", "group": "speeds"}}, "rulesets": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "functions": {}, "ungrouped_variables": ["d2farea", "area"], "statement": "

You are part of a team analysing a high speed car race. You are given the following graph mapping both the speed and acceleration 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 and acceleration in metres per second and metres per second squared respectively.

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

", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

A graph shows both the speed and acceleration of a car. Identify which line corresponds to which measurement, and calculate the acceleration during a portion of time.

"}}, {"name": " Using a speed graph to find distance", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Bradley Bush", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1521/"}], "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"}, "rulesets": {}, "type": "question", "ungrouped_variables": ["d2farea", "area"], "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)

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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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 ($ms^{-1}$).

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

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

Distance travelled $=$ [[0]]metres

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

Distance travelled $=$ [[0]]metres

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

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.

Average speed $=$ [[0]] ms^-1

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Multiplication and division of upper and lower bounds.

", "licence": "Creative Commons Attribution 4.0 International"}, "ungrouped_variables": ["speed", "distance", "time", "atime", "person", "pronouns", "verbs"], "statement": "

{person['name']} is a keen runner. {capitalise(pronouns['they'])} run{verbs} at an average speed of {speed}km/h, rounded to the nearest integer.

", "variable_groups": [], "rulesets": {}, "type": "question", "advice": "

We're not certain about some of the measurements given in this question - we only know the rounded values. This means that the true value could be lower or higher than the given measurement.

We can find upper and lower bounds for the given measurements. Any values we go on to calculate will also be uncertain and have upper and lower bounds.

To find bounds for a given measurement, we divide the degree of accuracy by 2 and subtract or add this to our estimate to get lower and upper bounds respectively.

For example, $52$ rounded to the nearest integer has a lower bound of $51.5$ and an upper bound of $52.5$.

a)

The distance travelled is given by

\\[ d = \\text{Average speed} \\times \\text{Time taken} \\]

We find bounds for speed and time first.

Lower bound for speed: $\\var{speed} - 0.5 = \\var{speed - 0.5} \\text{ km/h}$

Upper bound for speed: $\\var{speed} + 0.5 = \\var{speed + 0.5} \\text{ km/h}$

First, note that the speed is given in km/h and we want to find the distance in km. We will convert the given time into hours.

Lower bound for time taken:

\\begin{align} \\var{atime} - 0.5 &= \\var{atime - 0.5} \\text{ min} \\\\[0.5em]
&= \\frac{\\var{atime - 0.5}}{60} \\text{ h}
\\end{align}

Upper bound for time taken:

\\begin{align} \\var{atime} + 0.5 &= \\var{atime + 0.5} \\text{ min} \\\\[0.5em]
&= \\frac{\\var{atime + 0.5}}{60} \\text{ h}
\\end{align}

Since we're multiplying the speed and time together, the lower bound for distance is the slowest speed multiplied by the shortest time:

\\begin{align}
\\text{Lower bound} &= \\text{lower bound for speed} \\times \\text{lower bound for time}\\\\
&= \\var{speed - 0.5} \\times \\frac{\\var{(atime - 0.5)}}{60} \\\\
&= \\var{precround((speed-0.5)*(atime - 0.5)/60, 2 )} \\text{ km} \\quad \\text{(rounded to 2 decimal places).}
\\end{align}

The upper bound for distance is the fastest speed multiplied by the longest time:

\\begin{align}
\\text{Upper bound} &= \\text{upper bound for speed} \\times \\text{upper bound for time} \\\\
&= \\var{speed + 0.5} \\times \\frac{\\var{atime + 0.5}}{60} \\\\
&= \\var{precround((speed+0.5)*(atime + 0.5)/60, 2 )} \\text{ km} \\quad \\text{(rounded to 2 decimal places).}
\\end{align}

Hence,

\\[\\var{precround((speed-0.5)*(atime - 0.5)/60, 2 )} \\leq d \\lt \\var{precround((speed+0.5)*(atime + 0.5)/60, 2 )} \\text{.}\\]

b)

We're told the speed and the distance travelled, so the time taken is given by

\\[ t = \\frac{\\text{Distance travelled}}{\\text{Average speed}} \\]

We found upper and lower bounds for {person['name']}'s average speed above.

The distance of the evening run is given to the nearest kilometre, so we can compute bounds as follows:

Lower bound for distance: $\\var{distance} - 0.5 = \\var{distance - 0.5} \\mathrm{km}$

Upper bound for distance: $\\var{distance} + 0.5 = \\var{distance + 0.5} \\mathrm{km}$

The upper bound for the time taken is the longest distance divided by the slowest speed:

\\begin{align}
\\text{Upper bound} &= \\text{upper bound for distance} \\div \\text{lower bound for speed} \\\\
&= \\var{distance + 0.5} \\div \\var{speed - 0.5} \\\\
&= \\var{(distance + 0.5)/(speed - 0.5)} \\text{ hours.}
\\end{align}

We're asked for the answer in minutes, to two decimal places.

\\begin{align}
\\var{(distance + 0.5)/(speed - 0.5)} \\text{ hours} &= \\var{(distance + 0.5)/(speed - 0.5)}\\times 60 \\text{ min} \\\\
&= \\var{precround((distance + 0.5)/(speed - 0.5)*60, 2)} \\text{ minutes} \\quad \\text{(rounded to 2 decimal places)}
\\end{align}

The lower bound for time is the shortest distance divided by the fastest speed:

\\begin{align}
\\text{Lower bound} &= \\text{lower bound for distance} \\div \\text{upper bound for speed} \\\\
&= \\var{distance - 0.5} \\div \\var{speed + 0.5} \\\\
&= \\var{(distance - 0.5)/(speed + 0.5)} \\text{ hours.}
\\end{align}

Converting into minutes, to two decimal places:

\\begin{align}
\\var{(distance - 0.5)/(speed + 0.5)} \\text{ hours} &= \\var{(distance - 0.5)/(speed + 0.5)}\\times 60 \\text{ min} \\\\
&= \\var{precround((distance - 0.5)/(speed + 0.5)*60, 2)} \\text{ min.}
\\end{align}

Therefore, we cannot confidently say {person['name']}'s time was less than {time*60 +1} minutes as the upper bound for {pronouns['their']} time, $\\var{precround((distance + 0.5)/(speed - 0.5)*60, 2)}$ minutes, is above this threshold.

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We're not certain about some of the measurements given in this question - we only know the rounded values. This means that the true value could be lower or higher than the given measurement.

Compute upper and lower bounds for {person['name']}'s average speed and the time spent running, then use those to find upper and lower bounds for the distance travelled.

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Suppose {person['name']} ran for {atime} minutes, rounded to the nearest minute.

Using the rounded figures for {pronouns['their']} average speed and time spent running, calculate upper and lower bounds for the distance, $d$, that {person['name']} ran.

Round your answers to two decimal places.

[[0]] $\\leq d \\lt$ [[1]]km

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Now consider {person['name']}'s evening run. {capitalise(pronouns['they'])} covered a distance of {precround(distance,0)}km, rounded to the nearest kilometre.

{capitalise(pronouns['their'])} friend's record time for the evening run is exactly {time*60 +1} minutes.

Can we confidently say that {person['name']} beat {pronouns['their']} friend's record?

First, calculate the lower and upper bounds for {person['name']}'s time, $t$.

Round your answers to two decimal places.

[[0]] $\\leq t \\lt $ [[1]] minutes

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Therefore, we [[0]] confidently say {pronouns['their']} time was less than {time*60+1} minutes as the [[1]] bound for time is [[2]] the given threshold of {time*60+1} mins.

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can

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cannot

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upper

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lower

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above

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below

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equal to

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Distance in kms.

", "definition": "speed*time", "group": "Ungrouped variables", "name": "distance", "templateType": "anything"}, "atime": {"description": "

Time in minutes

", "definition": "random(5..30 #5)", "group": "Ungrouped variables", "name": "atime", "templateType": "anything"}, "verbs": {"description": "", "definition": "if(person['gender']='neutral','','s')", "group": "Ungrouped variables", "name": "verbs", "templateType": "anything"}, "person": {"description": "", "definition": "random_person()", "group": "Ungrouped variables", "name": "person", "templateType": "anything"}, "time": {"description": "

Time in hours.

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Speed in km/h.

", "definition": "random(7..11)", "group": "Ungrouped variables", "name": "speed", "templateType": "anything"}}, "variablesTest": {"maxRuns": "1000", "condition": ""}}, {"name": "Using compound units - speed", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "variable_groups": [], "preamble": {"js": "", "css": ""}, "type": "question", "parts": [{"variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "variableReplacements": [], "precisionMessage": "

Round your answer to $2$ decimal places.

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What was the runner's average speed, in metres per second?

[[0]] m/s

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Round your answer to $2$ decimal places.

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How fast is this in kilometres per hour?

[[0]] km/h

"}], "advice": "

a)

To find the average speed of the runner in meters per second (m/s), we divide the distance covered by the runner (in metres) by the time taken for the runner to run this distance (in seconds).

\\[
\\begin{align}
\\text{Average speed} &= \\displaystyle\\frac{\\var{distance}}{\\var{seconds}}\\\\
&= \\var{distance/seconds}\\\\
&= \\var{dpformat(distance/seconds,2)}\\; \\text{m/s} \\; (\\text{rounded to $2$ decimal places}).
\\end{align}
\\]

b)

We can convert the average speed of the runner that we calculated in a) in metres per second to kilometres per hour using the following two equivalences:

\\[1\\text{m} = \\displaystyle\\frac{1}{1000}\\text{km},\\]

\\[
1 \\; \\text{second} = \\displaystyle\\frac{1}{60} \\; \\text{minutes} = \\displaystyle\\frac{1}{3600} \\; \\text{hours}.
\\]

We know from a) that the average speed of the runner in m/s was $\\var{dpformat(distance/seconds,5)}$ m/s ($5$ d.p), so to convert this speed to km/h we first need to convert metres to kilometres,

\\[\\var{dpformat(distance/seconds,5)} \\; \\text{m/s} = \\var{dpformat(distance/seconds/1000,5)} \\text{km/s} \\; (5 \\; \\text{d.p})\\]

Then we convert seconds to hours,

\\[1 \\; \\text{second} = \\displaystyle\\frac{1}{3600} \\; \\text{hours}.\\]

Now we have

\\[\\var{dpformat(distance/seconds,5)} \\; \\text{m/s} = \\var{sigformat(distance/seconds/1000,5)} \\; \\text{kilometres per $\\displaystyle\\frac{1}{3600}$ hours}.\\]

We want a rate per one hour, so we multiply by $3600$ to obtain a measurement in km/h:

\\[\\begin{align}\\var{sigformat(distance/seconds/1000,5)} \\; \\text{kilometres per $\\displaystyle\\frac{1}{3600}$ hours} &= \\var{siground(distance/seconds/1000,5)*3600} \\; \\text{km}/\\text{h}\\\\&=\\var{dpformat(distance/seconds*3.6, 2)} \\; \\text{km}/\\text{h} \\; (\\text{rounded to $2$ decimal places}).\\end{align}\\]

\\[\\var{sigformat(distance/seconds/1000,5)} \\; \\text{kilometres per $\\displaystyle\\frac{1}{3600}$ hours} = \\var{distance/seconds*3.6} \\; \\text{km}/\\text{h}.\\]

Note that throughout this calculation we have rounded all figures to $5$ decimal places for convenience; when doing calculations which involve long decimals, you should always input the full figure into your calculator to avoid getting an incorrect answer due to rounding.

", "tags": ["taxonomy"], "variables": {"speed": {"templateType": "anything", "description": "

Athlete's speed, in m/s.

4m/s is about 9 mph, a bit faster than a jog. The current world record is 12m/s.

", "definition": "random(4..8#0)", "name": "speed", "group": "Ungrouped variables"}, "distance": {"templateType": "anything", "description": "

Distance that the runner ran.

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Time taken to cover the distance, in seconds.

", "definition": "floor(distance/speed)", "name": "seconds", "group": "Ungrouped variables"}}, "rulesets": {}, "variablesTest": {"condition": "", "maxRuns": "100"}, "functions": {}, "ungrouped_variables": ["distance", "seconds", "speed"], "statement": "

An athlete runs $\\var{distance}$ m in $\\var{seconds}$ seconds.

Round each of your answers to two decimal places.

", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Calculate a speed in m/s given distance and time taken, then convert that to km/hour

"}}, {"name": "Use speed and distance to calculate time ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "metadata": {"description": "

Calculate the time taken for a certain distance to be travelled given the average speed and the distance travelled.

Small, simple question.

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At a greyhound race, a fake rabbit moves around the inside of the track to motivate the dogs to run.

The track is $\\var{kilometres}$km long and the rabbit moves at a constant speed of $\\var{speed}$m/s.

", "advice": "

We are told that the track is $\\var{kilometres}$km long and that the speed of the rabbit is $\\var{speed}$m/s (metres per second).

Most people remember the relationship between speed, distance and time with the formula

\\[\\text{Average speed} = \\frac{\\text{Distance travelled}}{\\text{Total time taken}}.\\]

We can rearrange the formula for average speed to give us the formula for the time taken:

\\[\\text{Total time taken} = \\displaystyle\\frac{\\text{Distance travelled}}{\\text{Average speed}}.\\]

Firstly, we must convert the units of the length of the race from kilometres to metres. We know that $1\\text{km} = 1000\\text{m}$, therefore

\\begin{align}
\\var{kilometres}\\text{km} &= (\\var{kilometres} \\times 1000)\\text{m}\\\\
&= \\var{{kilometres}*1000}\\text{m}.
\\end{align}

Therefore, the time taken for the rabbit to finish one lap is

\\[\\displaystyle\\frac{\\var{{kilometres}*1000}}{\\var{speed}} = \\var{({kilometres}*1000)/{speed}} \\; \\text{seconds}.\\]

\\[ \\begin{align} \\displaystyle\\frac{\\var{{kilometres}*1000}}{\\var{speed}} &= \\var{({kilometres}*1000)/{speed}} \\; \\text{seconds}\\\\ &= \\var{dpformat(({kilometres}*1000)/{speed}, 0)} \\; \\text{seconds} \\; (\\text{to the nearest second}). \\end{align}\\]

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Speed, distance and time are related by the equation

\\[\\text{Average speed} = \\frac{\\text{Distance travelled}}{\\text{Total time taken}}.\\]

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Calculate the time taken for the rabbit to complete a circuit of the track.

[[0]] seconds Round your answer to the nearest second.

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Round your answer to the nearest second.

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Speed of the greyhound in metres per second.

", "group": "Ungrouped variables", "definition": "random(14..18)", "name": "speed", "templateType": "anything"}, "kilometres": {"description": "

Kilometres to be ran by the greyhound

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Calculations and graphs involving measurements of speed.

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