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Either paper 3 marks
Reuben cycles along a straight road at a steady speed.
The distance-time graph shows the first \(30\) seconds of the journey.
(a)How far had Reuben travelled after \(5\) seconds?
(b)Work out Reuben's speed. Give your answer in metres per second.
Show answer
(a) \(15\ \text{m}\)
(b) \(3\ \text{m/s}\)Find \(5\) on the time axis, go straight up until you meet the line, then straight across to the distance axis.
The distance axis is numbered every \(25\) m and ruled every \(5\) m in between, so the reading is exact.
The line meets the axis at \(15\), so Reuben had travelled \(15\) m.
On a distance-time graph the speed is the gradient of the line.
Part (a) gives a point on the line: after \(5\) s the distance is \(15\) m.
\(\text{speed} = \frac{ 15 }{ 5 } = 3\) m/s
- (b) M1distance \(15\) m over time \(5\) s (gradient)
- (b) A1speed \(3\) m/s
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Either paper 3 marks
Amara drives a delivery van away from the depot, stops to make a delivery, then drives further on to a second delivery.
(a)For how many minutes was the van stopped at the first delivery?
(b)How far was the van from the depot at the end of the journey?
(c)Between \(40\) minutes and \(50\) minutes the graph is horizontal.
Write down what this tells you about the van.
Show answer
(a) \(10\ \text{minutes}\)
(b) \(16\ \text{km}\)
(c) The van was not moving. It stayed 8 km from the depot while the delivery was made.The flat section runs from \(40\) minutes to \(50\) minutes.
\(50 - 40 = 10\) minutes
Read the height of the right-hand end of the graph across to the distance axis.
The van finishes \(16\) km from the depot.
The height of the graph is the distance from the depot. A horizontal section means that distance is not changing.
So the van was stationary, still \(8\) km from the depot, for those \(10\) minutes.
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Either paper 5 marks
Erin drives from home to a garden centre, stays there for a while, then drives straight home again.
(a)For how many minutes did Erin stay at the garden centre?
(b)Work out Erin's speed on the journey to the garden centre.
Give your answer in kilometres per hour.
(c)The journey there and the journey home cover the same distance.
Which of the two was faster?
Show answer
(a) \(10\ \text{minutes}\)
(b) \(45\ \text{km/h}\)
(c) The outward journeyThe middle section of the graph is horizontal, so the distance from home is not changing and Erin is not moving.
It runs from \(40\) minutes to \(50\) minutes:
\(50 - 40 = 10\) minutes
The outward part of the graph rises \(30\) km in \(40\) minutes.
There are \(60\) minutes in an hour, so multiply the distance per minute by \(60\):
\(\text{speed} = \frac{ 30 }{ 40 } \times 60 = 45\) km/h
Both parts of the journey cover \(30\) km. One takes \(40\) minutes and the other takes \(50\) minutes.
The same distance in less time means a greater speed, so the outward journey was faster.
On the graph that is the steeper of the two sloping sections: the speed out was \(45\) km/h and the speed home was \(36\) km/h.
- (b) M1distance divided by time, \(30\) km in \(40\) minutes, with minutes changed to hours
- (b) A1speed \(45\) km/h
- (c) M1same distance in less time means the greater speed: \(40\) minutes against \(50\) minutes
- (c) A1The outward journey — speed out \(45\) km/h, speed home \(36\) km/h
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Either paper 4 marks
A train travels from Derby to Nottingham at a steady speed.
The distance axis is numbered every \(20\) km, with a smaller square every \(5\) km.
(a)Use the graph to work out how far the train had travelled after \(30\) minutes.
(b)Work out the average speed of the train.
Give your answer in kilometres per hour.
Show answer
(a) \(50\ \text{km}\)
(b) \(100\ \text{km/h}\)Go up from \(30\) minutes to the line, then across to the distance axis.
The reading is not on a numbered line. It sits exactly halfway between \(40\) and \(60\), and half of \(20\) is \(10\):
\(40 + 10 = 50\) km
The right-hand end of the line is at \(100\) km after \(60\) minutes.
\(\text{speed} = \frac{ 100 }{ 60 } \times 60 = 100\) km/h
- (a) M1reading halfway between \(40\) km and \(60\) km
- (a) A1distance \(50\) km
- (b) M1\(\frac{ 100 }{ 60 } \times 60\), distance over time with minutes changed to hours
- (b) A1speed \(100\) km/h
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Either paper 5 marks
Two runners set off together and each runs at a steady speed for \(40\) seconds.
Runner \(X\) and runner \(Y\) are shown on the same grid.
(a)Write down which of the two runners was running faster.
(b)Find the speed of runner \(X\). Give your answer in metres per second.
(c)How much further had runner \(X\) run than runner \(Y\) after \(40\) seconds?
Show answer
(a) Runner X
(b) \(5\ \text{m/s}\)
(c) \(120\ \text{m}\)The steeper line is the faster runner, because it covers more distance in the same time.
Runner \(X\) has the steeper line.
Runner \(X\) covers \(200\) m in \(40\) s, so the gradient of that line is
\(\text{speed} = \frac{ 200 }{ 40 } = 5\) m/s
Read the two heights at \(40\) seconds: \(200\) m and \(80\) m.
The difference is \(120\) m.
- (b) M1distance \(200\) m divided by time \(40\) s
- (b) A1speed \(5\) m/s
- (c) M1heights \(200\) m and \(80\) m read at \(40\) s
- (c) A1difference \(120\) m
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Either paper 2 marks
Tomas's home is 800 metres from a shop.
Tomas walked from home to the shop at a constant speed in 11 minutes. Tomas stayed at the shop for 5 minutes. Tomas walked home at a constant speed in 11 minutes.
Tomas drew this distance-time graph to represent his journey.
Make two criticisms of his graph.
Criticism 1 ......................................................
Criticism 2 ......................................................
Show answer
Criticism 1: The stay at the shop is drawn going from 800 m to 1000 m, but it should be horizontal because nobody moves during a stay. Criticism 2: The walk home is drawn ending 1200 m from home, but it should end at home, 0 m from home.Check each section of the graph against the matching sentence of the story.
- Walk to the shop: the story says \(800\) m in \(11\) minutes. The graph reaches \(800\) m at \(11\) minutes. - Stay: the story says \(5\) minutes at the shop, so the graph should be horizontal from \(11\) to \(16\) minutes. On the graph it goes from \(800\) m at \(11\) minutes to \(1000\) m at \(16\) minutes. - Walk home: the story says \(11\) minutes, ending at \(0\) m. The graph takes \(11\) minutes and ends at \(1200\) m.
The graph disagrees with the story on the stay at the shop and on the walk home.
The stay at the shop is drawn going from 800 m to 1000 m, but it should be horizontal because nobody moves during a stay.
The walk home is drawn ending 1200 m from home, but it should end at home, 0 m from home.
- B1a criticism of the stay at the shop: The stay at the shop is drawn going from 800 m to 1000 m, but it should be horizontal because nobody moves during a stay.
- B1a criticism of the walk home: The walk home is drawn ending 1200 m from home, but it should end at home, 0 m from home.
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Either paper 4 marks
A tank contains \(40\) litres of water.
Water leaks out of the tank at a rate of \(0.8\) litres per minute.
The leak is stopped after \(20\) minutes.
(a)Show that, when the leak is stopped, the tank contains \(24\) litres of water.
(b)The tank is then refilled with water from a tap.
The graph shows the volume of water in the tank after the leak is stopped.
Complete this report by writing a number in each answer space.
The tap starts to refill the tank .......... minutes after the leak is stopped.
(c)The rate at which the tank refills is .......... litres per minute.
Show answer
(a) \(0.8 \times 20 = 16\) litres leaked, and \(40 - 16 = 24\) litres left
(b) \(7\ \text{minutes}\)
(c) \(2.5\ \text{litres per minute}\)The water lost is the rate multiplied by the time:
\(0.8 \times 20 = 16\) litres
Take this away from the water that was in the tank to begin with:
\(40 - 16 = 24\) litres
So the tank contains \(24\) litres when the leak is stopped.
While the line is level the volume is not changing, so the tap is not yet running.
The line starts to rise at \(7\) minutes.
The rate is the gradient of the rising part of the line. It rises from \(24\) litres to \(44\) litres, which is \(20\) litres, while the time goes from \(7\) minutes to \(15\) minutes, which is \(8\) minutes:
\(\frac{ 20 }{ 8 } = 2.5\) litres per minute
- (a) M1water lost \(0.8 \times 20 = 16\) litres
- (a) A1subtracting from the start, \(40 - 16 = 24\) litres, with units
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Either paper 4 marks
A taxi firm charges a fixed amount for every journey, plus an extra amount for each mile travelled.
The graph shows the cost of a journey of up to \(16\) miles.
(a)Write down the amount the taxi firm charges before it has travelled any distance at all.
(b)Find the gradient of the line.
(c)Write down what the gradient tells you about the cost of a journey.
Show answer
(a) \(\text{£}5\)
(b) \(2\)
(c) The taxi firm charges £2 for each extra mile travelled.A journey of \(0\) miles is where the line crosses the cost axis.
It crosses at \(5\), so the fixed charge is £\(5\).
Take the two ends of the line: \((0, 5)\) and \((16, 37)\).
\(\text{gradient} = \frac{ 37 - 5 }{ 16 - 0 } = \frac{ 32 }{ 16 } = 2\)
The gradient is how much the cost goes up for each extra mile, so it is measured in pounds per mile.
A gradient of \(2\) means the firm charges £\(2\) for every extra mile travelled — on top of the £\(5\) fixed charge.
- (b) M1change in cost \(32\) over change in distance \(16\)
- (b) A1gradient \(2\)
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Either paper 3 marks
Reuben visits a library.
He cycles to the library in 90 minutes at a speed of 9 miles per hour.
He stays at the library for 20 minutes.
He then cycles home.
The sketch graph represents the visit.
Work out the speed, in miles per hour, at which Reuben cycles home.
Give your answer to 1 decimal place.
Show answer
\(17.6\ \text{mph}\)The distance to the library is not on the graph, so find it from the outward journey.
\(90\) minutes is \(\frac{ 90 }{60}\) hours, so
\(\text{distance} = 9 \times \frac{ 90 }{60} = 13.5\) miles
The graph shows Reuben leaving at \(11:45\) and arriving home at \(14:21\), a total of \(156\) minutes.
The outward journey and the stay take \(90 + 20 = 110\) minutes, so the return journey takes
\(156 - 110 = 46\) minutes, which is \(\frac{ 46 }{60}\) hours.
\(\text{speed} = \frac{ 13.5 }{ 46 / 60 } = 13.5 \times \frac{60}{ 46 } \approx 17.6\) mph
- M1distance \(13.5\) miles from \(9 \times \frac{ 90 }{60}\)
- M1return time \(46\) minutes, from \(156 - 110\)
- A1speed \(17.6\) mph
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Either paper 4 marks
Rosa goes on a car journey from home.
For the first 30 minutes her average speed is 30 miles per hour.
Rosa then stops for 20 minutes.
Rosa then completes the journey at an average speed of 30 miles per hour.
The total journey time is 120 minutes.
(a)On the grid, draw a distance-time graph for Rosa's journey.
Plot the point where the journey starts, each point where the graph changes direction and the point where the journey ends, then join them.
(b)Write down the average speed for the whole journey.
You cannot draw on this page. Draw (a) on paper, then show the answer and check your drawing against the model.
Show answer
(a) Straight lines joining \((0, 0)\), \((30, 15)\), \((50, 15)\) and \((120, 50)\)
(b) \(25\ \text{mph}\)(a)First stage: \(30\) minutes is \(\frac{ 30 }{60}\) of an hour, so the distance is
\(30 \times \frac{ 30 }{60} = 15\) miles
The first point is \((30, 15)\).
During the stop the distance does not change, so the graph goes straight across to \((50, 15)\).
The last stage lasts
\(120 - 50 = 70\) minutes
and covers
\(30 \times \frac{ 70 }{60} = 35\) miles.
The total distance is \(15 + 35 = 50\) miles, so the last point is \((120, 50)\).
The whole journey is \(50\) miles in \(120\) minutes, and the stop counts.
\(\text{average speed} = 50 \div \frac{ 120 }{60} = 25\) mph
- (a) B1at least two of the four points plotted, for example \((0, 0)\) and \((30, 15)\): \(30 \times \frac{ 30 }{60} = 15\) miles
- (a) B1at least three of the four points plotted: the stop ends at \((50, 15)\), the same distance \(20\) minutes later
- (a) B1all four points plotted and joined, ending at \((120, 50)\): last stage \(70\) minutes, \(35\) miles added