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Exam questions · Maths · Graphs

Real-Life Graphs

  • 6 exam questions
  • 20 marks
  • 9 quick checks
  1. 1 Work out [6 marks]

    The graph shows the velocity of a car during a 10 second journey. (a) Work out the acceleration of the car in the first 2 seconds. [2 marks] (b) Work out the deceleration of the car in the last 4 seconds. [2 marks] (c) Work out the total distance travelled by the car. [2 marks]

    A velocity-time graph that speeds up to 8 m/s in 2 seconds, stays constant until 6 seconds and slows to rest at 10 seconds.
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    Model answer

    (a) \(\dfrac{8}{2} = 4\) m/s\(^2\). (b) \(\dfrac{8}{4} = 2\) m/s\(^2\). (c) The shape is a trapezium with parallel sides 10 and 4 and height 8, so the distance is \(\dfrac{1}{2}(10 + 4) \times 8 = 56\) m.

    Mark scheme

    • (a) \(\dfrac{8}{2}\) — M1
    • (a) 4 m/s\(^2\) — A1
    • (b) \(\dfrac{8}{4}\) — M1
    • (b) 2 m/s\(^2\) — A1
    • (c) \(\dfrac{1}{2}(10 + 4) \times 8\) or the areas of the three parts added — M1
    • (c) 56 m — A1
  2. 2 Work out [3 marks]

    A mobile phone plan costs \(\pounds 12\) a month plus 4p for each minute of calls. (a) Work out the cost of a month with 150 minutes of calls. [2 marks] (b) Write down what the number 12 would represent on a graph of cost against minutes. [1 mark]

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    Model answer

    (a) \(150 \times 0.04 = 6\), so the cost is \(12 + 6 = \pounds 18\). (b) It is the \(y\)-intercept, the fixed monthly charge.

    Mark scheme

    • (a) \(150 \times 0.04\) or \(150 \times 4 = 600\) — M1
    • (a) \(\pounds 18\) — A1
    • (b) The fixed charge, or the cost with no calls — B1
  3. 3 Work out [2 marks]

    The temperature \(F\) in degrees Fahrenheit is given by \(F = 1.8C + 32\), where \(C\) is the temperature in degrees Celsius. Work out \(F\) when \(C = 20\).

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    Model answer

    \(F = 1.8 \times 20 + 32 = 36 + 32 = 68\).

    Mark scheme

    • \(1.8 \times 20\) or 36 seen — M1
    • 68 — A1
  4. 4 Work out [4 marks]

    A cyclist starts from rest and speeds up steadily to 6 m/s in 3 seconds. She then cycles at 6 m/s for 10 seconds, and then slows down steadily to rest in 3 seconds. (a) Work out her acceleration in the first 3 seconds. [1 mark] (b) Work out the total distance she travels. [3 marks]

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    Model answer

    (a) \(\dfrac{6}{3} = 2\) m/s\(^2\). (b) The two triangles are each \(\dfrac{1}{2} \times 3 \times 6 = 9\) and the rectangle is \(10 \times 6 = 60\), so the total is \(9 + 60 + 9 = 78\) m.

    Mark scheme

    • (a) 2 m/s\(^2\) — B1
    • (b) \(\dfrac{1}{2} \times 3 \times 6\) or \(10 \times 6\) — M1
    • (b) \(9 + 60 + 9\) or an equivalent total — M1
    • (b) 78 m — A1
  5. 5 Explain [2 marks]

    Water is poured at a steady rate into a glass that is wider at the top than at the bottom. Describe the graph of the depth of the water against time.

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    Model answer

    The depth rises quickly at first, because the bottom is narrow, and then more slowly as the glass gets wider. So the graph starts steep and gets less steep.

    Mark scheme

    • Starts steep — Q1
    • Gets less steep, or flatter, as the glass fills — Q1
  6. 6 Work out [3 marks]

    The velocity of a car is measured every 2 seconds. At times \(t = 0, 2, 4, 6\) seconds the velocity is \(v = 0, 6, 10, 12\) metres per second. Use trapezia to estimate the distance travelled in the 6 seconds.

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    Model answer

    The areas of the three trapezia are \(\dfrac{1}{2} \times 2 \times (0 + 6) = 6\), \(\dfrac{1}{2} \times 2 \times (6 + 10) = 16\) and \(\dfrac{1}{2} \times 2 \times (10 + 12) = 22\). The total is \(6 + 16 + 22 = 44\) m.

    Mark scheme

    • One trapezium area found correctly — M1
    • \(6 + 16 + 22\) or equivalent — M1
    • 44 m — A1

Quick check

  1. 1

    On a conversion graph, 5 miles is about 8 km. About how many kilometres is 30 miles?

    1. A150 km
    2. B48 km
    3. C38 km
    4. D18.75 km
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    B: 48 km

    30 miles is 6 lots of 5 miles, so \(6 \times 8 = 48\) km.

  2. 2

    What does the gradient of a velocity-time graph show?

    1. AAcceleration
    2. BDistance travelled
    3. CTime taken
    4. DTotal speed
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    A: Acceleration

    Gradient is change in velocity divided by time, which is acceleration.

  3. 3

    What does the area under a velocity-time graph show?

    1. AAcceleration
    2. BSpeed
    3. CTime taken
    4. DDistance travelled
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    D: Distance travelled

    Velocity multiplied by time gives distance.

  4. 4

    A car speeds up from 0 to 12 m/s in 4 seconds. What is its acceleration?

    1. A48 m/s\(^2\)
    2. B8 m/s\(^2\)
    3. C3 m/s\(^2\)
    4. D12 m/s\(^2\)
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    C: 3 m/s\(^2\)

    \(\dfrac{12}{4} = 3\).

  5. 5

    A taxi costs \(\pounds 3\) plus \(\pounds 2\) for each kilometre. What is the cost of a 7 km journey?

    1. A\(\pounds 14\)
    2. B\(\pounds 17\)
    3. C\(\pounds 21\)
    4. D\(\pounds 10\)
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    B: \(\pounds 17\)

    \(3 + 2 \times 7 = 17\).

  6. 6

    On a graph of taxi cost against distance, what does the \(y\)-intercept mean?

    1. AThe fixed starting charge
    2. BThe cost per kilometre
    3. CThe total cost
    4. DThe longest journey
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    A: The fixed starting charge

    The intercept is the cost for 0 km.

  7. 7

    A container gets wider towards the top and is filled at a steady rate. What happens to the depth-time graph?

    1. AIt is a straight line
    2. BIt gets steeper and steeper
    3. CIt is horizontal
    4. DIt rises more and more slowly, so it flattens
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    D: It rises more and more slowly, so it flattens

    The wider the container, the more slowly the depth rises.

  8. 8

    A velocity-time graph is a triangle that rises from 0 to 8 m/s in 5 seconds. What distance does it show?

    1. A40 m
    2. B13 m
    3. C20 m
    4. D1.6 m
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    C: 20 m

    Area \(= \dfrac{1}{2} \times 5 \times 8 = 20\).

  9. 9

    How can you estimate the speed at one moment from a curved distance-time graph?

    1. AFind the area under the curve
    2. BDraw a tangent and find its gradient
    3. CRead the highest point
    4. DRead the value at the end
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    B: Draw a tangent and find its gradient

    The gradient of the tangent is the rate of change at that point.