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Exam questions · Maths · Graphs

Real-Life Graphs

  • 6 exam questions
  • 22 marks
  • 9 quick checks
  1. 1 Calculate [7 marks]

    The graph shows the velocity of a train during a 20 second journey. (a) Calculate the acceleration of the train in the first 5 seconds. [2 marks] (b) Calculate the deceleration of the train in the last 5 seconds. [2 marks] (c) Calculate the total distance travelled by the train. [3 marks]

    A velocity-time graph that speeds up to 10 m/s in 5 seconds, stays constant until 15 seconds and slows to 4 m/s at 20 seconds.
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    Model answer

    (a) \(\dfrac{10}{5} = 2\) m/s\(^2\). (b) The velocity falls from 10 to 4 in 5 seconds, so \(\dfrac{10 - 4}{5} = 1.2\) m/s\(^2\). (c) The areas are \(\dfrac{1}{2} \times 5 \times 10 = 25\), \(10 \times 10 = 100\) and \(\dfrac{1}{2}(10 + 4) \times 5 = 35\), so the total is \(25 + 100 + 35 = 160\) m.

    Mark scheme

    • (a) \(\dfrac{10}{5}\) — M1
    • (a) 2 m/s\(^2\) — A1
    • (b) \(\dfrac{10 - 4}{5}\) — M1
    • (b) 1.2 m/s\(^2\) — A1
    • (c) At least two of 25, 100, 35 found — M1
    • (c) \(25 + 100 + 35\) — M1
    • (c) 160 m — A1
  2. 2 Calculate [3 marks]

    A gym charges a joining fee of \(\pounds 20\) and \(\pounds 15\) for each month of membership. (a) Calculate the total cost after 6 months. [1 mark] (b) Write a formula for the total cost \(C\) pounds after \(m\) months. [2 marks]

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    Model answer

    (a) \(20 + 15 \times 6 = \pounds 110\). (b) \(C = 15m + 20\).

    Mark scheme

    • (a) \(\pounds 110\) — B1
    • (b) \(15m\) seen — M1
    • (b) \(C = 15m + 20\) — A1
  3. 3 Calculate [3 marks]

    A conversion graph from pounds to euros is a straight line through \((0, 0)\) and \((50, 60)\). (a) Calculate the gradient of the line. [2 marks] (b) What does the gradient represent? [1 mark]

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    Model answer

    (a) \(\dfrac{60}{50} = 1.2\). (b) It is the number of euros for each pound, so \(\pounds 1 = \euro 1.20\).

    Mark scheme

    • (a) \(\dfrac{60}{50}\) — M1
    • (a) 1.2 — A1
    • (b) The number of euros for one pound — C1
  4. 4 Calculate [4 marks]

    A car slows down steadily from 20 m/s to rest in 8 seconds. (a) Calculate the deceleration of the car. [2 marks] (b) Calculate the distance travelled while it slows down. [2 marks]

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    Model answer

    (a) \(\dfrac{20}{8} = 2.5\) m/s\(^2\). (b) The area of the triangle is \(\dfrac{1}{2} \times 8 \times 20 = 80\) m.

    Mark scheme

    • (a) \(\dfrac{20}{8}\) — M1
    • (a) 2.5 m/s\(^2\) — A1
    • (b) \(\dfrac{1}{2} \times 8 \times 20\) — M1
    • (b) 80 m — A1
  5. 5 Explain [2 marks]

    Water is poured at a steady rate into a bottle that is wide at the bottom and narrow at the top. Describe how the graph of depth against time changes as the bottle fills.

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    Model answer

    At first the bottle is wide, so the depth rises slowly and the graph is shallow. Near the top the bottle is narrow, so the depth rises quickly and the graph becomes steeper.

    Mark scheme

    • The graph starts shallow — C1
    • and gets steeper as the bottle fills — C1
  6. 6 Calculate [3 marks]

    The velocity of a runner is measured every 5 seconds. At times \(t = 0, 5, 10, 15\) seconds the velocity is \(v = 0, 8, 10, 4\) metres per second. Use trapezia to estimate the distance run in the 15 seconds.

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    Model answer

    The areas of the three trapezia are \(\dfrac{1}{2} \times 5 \times 8 = 20\), \(\dfrac{1}{2} \times 5 \times (8 + 10) = 45\) and \(\dfrac{1}{2} \times 5 \times (10 + 4) = 35\). The total is \(20 + 45 + 35 = 100\) m.

    Mark scheme

    • One trapezium area found correctly — M1
    • \(20 + 45 + 35\) or equivalent — M1
    • 100 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.