Exam questions · Maths
Graphs
- 30 exam questions
- 94 marks
- 45 quick checks
Straight-Line Graphs
Just this lesson-
1 Calculate [2 marks]
Calculate the gradient of the line through \((1, 2)\) and \((5, 10)\).
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Model answer
\(\dfrac{10 - 2}{5 - 1} = \dfrac{8}{4} = 2\).
Mark scheme
- \(\dfrac{10 - 2}{5 - 1}\) — M1
- 2 — A1
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2 Complete [3 marks]
(a) Complete the table of values for \(y = 4x - 3\). \(x = 0, 1, 2, 3\) [2 marks] (b) Write down the \(y\)-intercept of the line \(y = 4x - 3\). [1 mark]
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Model answer
(a) The values are \(-3, 1, 5, 9\). (b) The \(y\)-intercept is \(-3\), where \(x = 0\).
Mark scheme
- (a) At least two correct values — M1
- (a) \(-3, 1, 5, 9\) — A1
- (b) \(-3\) — B1
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3 Calculate [4 marks]
The diagram shows a straight line passing through the points \(P\) and \(Q\). (a) Calculate the gradient of the line. [2 marks] (b) Write down the equation of the line. [2 marks]
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Model answer
(a) \(\dfrac{7 - 1}{3 - 1} = \dfrac{6}{2} = 3\). (b) The line crosses the \(y\)-axis at \(-2\), so \(y = 3x - 2\).
Mark scheme
- (a) \(\dfrac{7 - 1}{3 - 1}\) — M1
- (a) 3 — A1
- (b) \(y = 3x + c\) or \(y = mx - 2\) — M1
- (b) \(y = 3x - 2\) — A1
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4 Find [2 marks]
Find the equation of the line that is parallel to \(y = 3x - 7\) and passes through the point \((0, 2)\).
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Model answer
The gradient is 3 and the \(y\)-intercept is 2, so \(y = 3x + 2\).
Mark scheme
- Gradient 3 or \(c = 2\) used — M1
- \(y = 3x + 2\) — A1
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5 Find [3 marks]
A straight line has gradient \(-2\) and passes through the point \((0, 5)\). (a) Write down the equation of the line. [1 mark] (b) Does the point \((4, -3)\) lie on the line? Show how you know. [2 marks]
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Model answer
(a) \(y = -2x + 5\). (b) When \(x = 4\), \(y = -2 \times 4 + 5 = -3\), so the point lies on the line.
Mark scheme
- (a) \(y = -2x + 5\) — B1
- (b) \(-2 \times 4 + 5\) or \(-8 + 5\) — M1
- (b) \(-3\) with a conclusion — A1
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6 Show that [4 marks]
\(A\) is the point \((-3, 2)\) and \(B\) is the point \((5, 6)\). (a) Work out the gradient of \(AB\). [2 marks] (b) Show that the line \(y = \dfrac{1}{2}x + 3.5\) passes through \(A\). [2 marks]
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Model answer
(a) \(\dfrac{6 - 2}{5 - (-3)} = \dfrac{4}{8} = \dfrac{1}{2}\). (b) When \(x = -3\), \(y = \dfrac{1}{2} \times (-3) + 3.5 = -1.5 + 3.5 = 2\), so the line passes through \(A\).
Mark scheme
- (a) \(\dfrac{6 - 2}{5 - (-3)}\) — M1
- (a) \(\dfrac{1}{2}\) — A1
- (b) \(\dfrac{1}{2} \times (-3) + 3.5\) — M1
- (b) 2 with a conclusion — A1
Quick check
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1
What is the equation of the \(x\)-axis?
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B: \(y = 0\)
Every point on the \(x\)-axis has \(y = 0\).
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2
Which point is on the line \(y = 3x - 2\)?
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A: \((4, 10)\)
Put in the coordinates: \(3 \times 4 - 2 = 10\), so \((4, 10)\) fits.
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3
Work out the gradient of the line through \((1, 7)\) and \((4, 1)\).
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D: \(-2\)
\(\dfrac{1 - 7}{4 - 1} = \dfrac{-6}{3} = -2\).
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4
Which line is parallel to \(y = 3x + 1\)?
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C: \(y = 3x - 5\)
Parallel lines have the same gradient, 3.
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5
What is the equation of the vertical line through 4 on the \(x\)-axis?
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B: \(x = 4\)
Every point on the line has \(x = 4\).
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6
What is the gradient of the line \(y = 5 - 3x\)?
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A: \(-3\)
Written as \(y = -3x + 5\), the number multiplying \(x\) is \(-3\).
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7
What is the value of \(y\) on the line \(y = 2x - 1\) when \(x = -1\)?
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D: \(-3\)
\(2 \times (-1) - 1 = -2 - 1 = -3\).
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8
Where does the line \(y = 3x + 2\) cross the \(y\)-axis?
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C: \((0, 2)\)
The number on its own, 2, is the \(y\)-intercept.
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9
Work out the gradient of the line through \((-2, 3)\) and \((4, -9)\).
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B: \(-2\)
\(\dfrac{-9 - 3}{4 - (-2)} = \dfrac{-12}{6} = -2\).
Equations of Straight Lines
Just this lesson-
1 Calculate [2 marks]
Calculate the midpoint of \((-3, -2)\) and \((5, 4)\).
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Model answer
\(\left(\dfrac{-3 + 5}{2}, \dfrac{-2 + 4}{2}\right) = (1, 1)\).
Mark scheme
- One coordinate correct — M1
- \((1, 1)\) — A1
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2 Calculate [5 marks]
The diagram shows a straight line through the points \(A\) and \(B\). (a) Calculate the gradient of \(AB\). [2 marks] (b) Find the equation of \(AB\). [2 marks] (c) Find the coordinates of the midpoint of \(AB\). [1 mark]
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Model answer
(a) \(\dfrac{0 - 6}{4 - 0} = -\dfrac{3}{2}\). (b) The line crosses the \(y\)-axis at 6, so \(y = -\dfrac{3}{2}x + 6\). (c) \(\left(\dfrac{0 + 4}{2}, \dfrac{6 + 0}{2}\right) = (2, 3)\).
Mark scheme
- (a) \(\dfrac{0 - 6}{4 - 0}\) — M1
- (a) \(-\dfrac{3}{2}\) — A1
- (b) \(y = -\dfrac{3}{2}x + c\) or \(y = mx + 6\) — M1
- (b) \(y = -\dfrac{3}{2}x + 6\) — A1
- (c) \((2, 3)\) — B1
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3 Find [3 marks]
Find the equation of the line that is parallel to \(y = 4x + 1\) and passes through \((1, 9)\).
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Model answer
The gradient is 4, so \(y = 4x + c\). Putting in \((1, 9)\) gives \(9 = 4 + c\), so \(c = 5\) and \(y = 4x + 5\).
Mark scheme
- \(y = 4x + c\) — M1
- \(9 = 4 \times 1 + c\) — M1
- \(y = 4x + 5\) — A1
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4 Find [3 marks]
A line has equation \(2x - 3y = 12\). Find the coordinates of the points where the line crosses the \(x\)-axis and the \(y\)-axis.
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Model answer
When \(y = 0\), \(2x = 12\) and \(x = 6\), so the point is \((6, 0)\). When \(x = 0\), \(-3y = 12\) and \(y = -4\), so the point is \((0, -4)\).
Mark scheme
- \(y = 0\) or \(x = 0\) used — M1
- \((6, 0)\) — A1
- \((0, -4)\) — A1
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5 Find [3 marks]
A line has equation \(3y + x = 6\). Find the gradient of a line that is perpendicular to it.
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Model answer
Rearranging, \(3y = -x + 6\), so \(y = -\dfrac{1}{3}x + 2\) and the gradient is \(-\dfrac{1}{3}\). The perpendicular gradient is 3.
Mark scheme
- \(y = -\dfrac{1}{3}x + 2\) or the gradient \(-\dfrac{1}{3}\) — M1
- Negative reciprocal used — M1
- 3 — A1
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6 Calculate [4 marks]
\(A\) is the point \((3, -2)\) and \(B\) is the point \((-2, 10)\). (a) Calculate the coordinates of the midpoint of \(AB\). [2 marks] (b) Calculate the length of \(AB\). [2 marks]
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Model answer
(a) \(\left(\dfrac{3 + (-2)}{2}, \dfrac{-2 + 10}{2}\right) = (0.5, 4)\). (b) The differences are 5 and 12, so \(AB = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\).
Mark scheme
- (a) One coordinate correct — M1
- (a) \((0.5, 4)\) — A1
- (b) \(\sqrt{5^2 + 12^2}\) — M1
- (b) 13 — A1
Quick check
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1
What is the equation of the line through \((2, 1)\) and \((6, 9)\)?
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C: \(y = 2x - 3\)
The gradient is \(\dfrac{8}{4} = 2\). Then \(1 = 2 \times 2 + c\) gives \(c = -3\).
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2
What is the midpoint of \((2, 1)\) and \((6, 9)\)?
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B: \((4, 5)\)
\(\left(\dfrac{2 + 6}{2}, \dfrac{1 + 9}{2}\right) = (4, 5)\).
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3
A line is parallel to \(y = 5x - 2\). What is its gradient?
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A: \(5\)
Parallel lines have equal gradients.
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4
A line has gradient 3 and passes through \((2, 9)\). What is its equation?
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D: \(y = 3x + 3\)
\(9 = 3 \times 2 + c\) gives \(c = 3\).
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5
What is the gradient of the line \(2y - 4x = 6\)?
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C: \(2\)
Rearranged, \(2y = 4x + 6\), so \(y = 2x + 3\).
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6
What is the midpoint of \((0, 4)\) and \((6, 10)\)?
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B: \((3, 7)\)
\(\left(\dfrac{0 + 6}{2}, \dfrac{4 + 10}{2}\right) = (3, 7)\).
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7
Where does the line \(3x + 2y = 12\) cross the axes?
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A: \((0, 6)\) and \((4, 0)\)
Put \(x = 0\) to get \(y = 6\), and \(y = 0\) to get \(x = 4\).
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8
What is the gradient of a line perpendicular to a line with gradient 4?
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D: \(-\dfrac{1}{4}\)
Perpendicular gradients multiply to \(-1\), so the new gradient is \(-\dfrac{1}{4}\).
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9
What is the equation of the line perpendicular to \(y = 2x + 3\) through \((4, 1)\)?
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C: \(y = -\dfrac{1}{2}x + 3\)
The gradient is \(-\dfrac{1}{2}\). Then \(1 = -2 + c\), so \(c = 3\).
Quadratic Graphs
Just this lesson-
1 Complete [2 marks]
Complete the table of values for \(y = x^2 - 2x\). \(x = -1, 0, 1, 2, 3\)
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Model answer
The values are \(3, 0, -1, 0, 3\).
Mark scheme
- At least three correct values — M1
- \(3, 0, -1, 0, 3\) — A1
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2 Write down [4 marks]
The diagram shows the graph of \(y = 4x - x^2\). (a) Write down the coordinates of the maximum point. [1 mark] (b) Use the graph to solve \(4x - x^2 = 0\). [2 marks] (c) Write down the equation of the line of symmetry. [1 mark]
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Model answer
(a) The highest point is \((2, 4)\). (b) The curve crosses the \(x\)-axis at \(x = 0\) and \(x = 4\). (c) The line of symmetry is \(x = 2\).
Mark scheme
- (a) \((2, 4)\) — B1
- (b) One of \(x = 0\) or \(x = 4\) — M1
- (b) \(x = 0\) and \(x = 4\) — A1
- (c) \(x = 2\) — B1
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3 Find [3 marks]
A curve has equation \(y = x^2 + 2x - 3\). (a) Write down the coordinates of the point where the curve crosses the \(y\)-axis. [1 mark] (b) Solve \(x^2 + 2x - 3 = 0\) to find where the curve crosses the \(x\)-axis. [2 marks]
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Model answer
(a) \((0, -3)\). (b) \(x^2 + 2x - 3 = (x + 3)(x - 1) = 0\), so \(x = -3\) and \(x = 1\).
Mark scheme
- (a) \((0, -3)\) — B1
- (b) \((x + 3)(x - 1)\) — M1
- (b) \(x = -3\) and \(x = 1\) — A1
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4 Find [3 marks]
The curve \(y = x^2 - 4x + k\) has its turning point at \((2, -1)\). Find the value of \(k\).
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Model answer
At the turning point \(x = 2\) and \(y = -1\), so \(-1 = 2^2 - 4 \times 2 + k = -4 + k\) and \(k = 3\).
Mark scheme
- \(x = 2\) and \(y = -1\) substituted — M1
- \(-1 = 4 - 8 + k\) — M1
- 3 — A1
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5 Calculate [3 marks]
(a) Write \(x^2 - 10x + 7\) in the form \((x - a)^2 + b\). [2 marks] (b) Write down the coordinates of the turning point of the graph of \(y = x^2 - 10x + 7\). [1 mark]
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Model answer
(a) \((x - 5)^2 - 25 + 7 = (x - 5)^2 - 18\). (b) The turning point is \((5, -18)\).
Mark scheme
- (a) \((x - 5)^2\) seen — M1
- (a) \((x - 5)^2 - 18\) — A1
- (b) \((5, -18)\) — B1
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6 Find [2 marks]
The graph of \(y = x^2 - 4x + 3\) crosses the \(x\)-axis at \(x = 1\) and \(x = 3\). Write down the values of \(x\) for which \(y\) is negative.
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Model answer
The graph is a U shape, so it is below the \(x\)-axis between the roots: \(1 < x < 3\).
Mark scheme
- Between the roots 1 and 3 identified — M1
- \(1 < x < 3\) — A1
Quick check
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1
What is the shape of the graph of a quadratic equation?
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D: A parabola, a smooth U or upside-down U
Quadratic graphs are parabolas.
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2
What are the roots of a graph?
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C: The \(x\)-values where the curve crosses the \(x\)-axis
At the roots \(y = 0\), so the curve meets the \(x\)-axis.
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3
Work out \(y\) when \(x = -2\) on \(y = x^2 - 2x - 3\).
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B: \(5\)
\((-2)^2 - 2 \times (-2) - 3 = 4 + 4 - 3 = 5\).
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4
What is the \(y\)-intercept of \(y = x^2 + 4x - 7\)?
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A: \(-7\)
Put \(x = 0\): \(y = -7\).
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5
The graph of \(y = x^2 - 2x - 3\) crosses the \(x\)-axis at \(-1\) and 3. What are the solutions of \(x^2 - 2x - 3 = 0\)?
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D: \(x = -1\) and \(x = 3\)
The solutions are the \(x\)-values where \(y = 0\).
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6
A parabola crosses the \(x\)-axis at \(x = 1\) and \(x = 5\). What is its line of symmetry?
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C: \(x = 3\)
The line of symmetry is halfway between the roots: \(\dfrac{1 + 5}{2} = 3\).
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7
The curve \(y = x^2 - 4x + 3\) crosses the \(x\)-axis at 1 and 3. What are the coordinates of its turning point?
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B: \((2, -1)\)
\(x = 2\) is halfway between the roots. Then \(y = 4 - 8 + 3 = -1\).
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8
Which line do you draw on the graph of \(y = x^2 - 2x - 3\) to solve \(x^2 - 2x - 3 = 2\)?
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A: \(y = 2\)
Solutions are where the curve meets the horizontal line \(y = 2\).
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9
What are the coordinates of the turning point of \(y = (x - 3)^2 - 4\)?
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D: \((3, -4)\)
In \(y = (x - a)^2 + b\) the turning point is \((a, b)\).
Cubic, Reciprocal and Other Graphs
Just this lesson-
1 Match [3 marks]
The diagram shows three graphs, \(A\), \(B\) and \(C\). The three equations are \(y = x^2\), \(y = 2x - 1\) and \(y = x^3 - 3x\). Match each equation to the correct graph. [3 marks]
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Model answer
Graph \(A\) is an S shape with a hill and a valley, so \(y = x^3 - 3x\). Graph \(B\) is a U shape touching the origin, so \(y = x^2\). Graph \(C\) is a straight line, so \(y = 2x - 1\).
Mark scheme
- \(A\) is \(y = x^3 - 3x\) — B1
- \(B\) is \(y = x^2\) — B1
- \(C\) is \(y = 2x - 1\) — B1
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2 Complete [2 marks]
Complete the table of values for \(y = x^3 - x\). \(x = -2, -1, 0, 1, 2\)
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Model answer
The values are \(-6, 0, 0, 0, 6\).
Mark scheme
- At least three correct values — M1
- \(-6, 0, 0, 0, 6\) — A1
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3 Complete [3 marks]
(a) Complete the table of values for \(y = \dfrac{12}{x}\). \(x = 1, 2, 3, 4, 6, 12\) [2 marks] (b) For which value of \(x\) is the graph of \(y = \dfrac{12}{x}\) not defined? [1 mark]
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Model answer
(a) The values are \(12, 6, 4, 3, 2, 1\). (b) It is not defined at \(x = 0\), because you cannot divide by zero.
Mark scheme
- (a) At least four correct values — M1
- (a) \(12, 6, 4, 3, 2, 1\) — A1
- (b) \(x = 0\) — B1
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4 Calculate [3 marks]
A population of bacteria is 100 at the start. It doubles every hour, so after \(t\) hours it is \(P = 100 \times 2^t\). (a) Calculate \(P\) when \(t = 3\). [1 mark] (b) Calculate \(P\) when \(t = 5\). [1 mark] (c) Explain why the graph of \(P\) against \(t\) never touches the \(t\)-axis. [1 mark]
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Model answer
(a) \(100 \times 8 = 800\). (b) \(100 \times 32 = 3200\). (c) \(2^t\) is always greater than 0, so \(P\) is never 0.
Mark scheme
- (a) 800 — B1
- (b) 3200 — B1
- (c) \(2^t\) is never zero, so the population is never 0 — C1
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5 Calculate [3 marks]
A curve has equation \(y = x^2 - 9\). (a) Calculate the coordinates of the points where the curve crosses the \(x\)-axis. [2 marks] (b) Write down the \(y\)-intercept. [1 mark]
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Model answer
(a) When \(y = 0\), \(x^2 = 9\), so \(x = 3\) or \(x = -3\). The points are \((3, 0)\) and \((-3, 0)\). (b) The \(y\)-intercept is \(-9\).
Mark scheme
- (a) \(x^2 = 9\) — M1
- (a) \((3, 0)\) and \((-3, 0)\) — A1
- (b) \(-9\) — B1
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6 Show that [3 marks]
A circle has equation \(x^2 + y^2 = 13\). (a) Write down the radius of the circle in surd form. [1 mark] (b) Show that the point \((2, 3)\) lies on the circle. [2 marks]
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Model answer
(a) The radius is \(\sqrt{13}\). (b) \(2^2 + 3^2 = 4 + 9 = 13\), so the point lies on the circle.
Mark scheme
- (a) \(\sqrt{13}\) — B1
- (b) \(2^2 + 3^2\) or \(4 + 9\) — M1
- (b) 13 with a conclusion — A1
Quick check
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1
What shape is the graph of \(y = x^3\)?
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A: An S-shaped curve through the origin
Cubic graphs have an S shape.
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2
What is special about the graph of \(y = \dfrac{1}{x}\)?
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D: It has two branches and never touches either axis
You cannot divide by 0, and \(\dfrac{1}{x}\) is never 0.
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3
Where does the graph of \(y = 3^x\) cross the \(y\)-axis?
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C: \((0, 1)\)
\(3^0 = 1\).
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4
What is the value of \(x^3\) when \(x = -3\)?
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B: \(-27\)
\((-3) \times (-3) \times (-3) = -27\).
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5
Which of these equations gives a cubic graph?
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A: \(y = x^3 + 1\)
A cubic has \(x^3\) as its highest power.
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6
Work out \(y\) when \(x = 2\) on \(y = x^3 - 3x\).
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D: \(2\)
\(8 - 6 = 2\).
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7
What is \(2^3\)?
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C: \(8\)
\(2 \times 2 \times 2 = 8\).
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8
What shape is the graph of \(y = -x^2\)?
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B: An upside-down U
A negative \(x^2\) term turns the parabola upside down.
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9
What is the radius of the circle \(x^2 + y^2 = 36\)?
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A: \(6\)
The radius is \(\sqrt{36} = 6\).
Real-Life Graphs
Just this lesson-
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]
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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
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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
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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
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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
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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
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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
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1
On a conversion graph, 5 miles is about 8 km. About how many kilometres is 30 miles?
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B: 48 km
30 miles is 6 lots of 5 miles, so \(6 \times 8 = 48\) km.
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2
What does the gradient of a velocity-time graph show?
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A: Acceleration
Gradient is change in velocity divided by time, which is acceleration.
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3
What does the area under a velocity-time graph show?
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D: Distance travelled
Velocity multiplied by time gives distance.
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4
A car speeds up from 0 to 12 m/s in 4 seconds. What is its acceleration?
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C: 3 m/s\(^2\)
\(\dfrac{12}{4} = 3\).
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5
A taxi costs \(\pounds 3\) plus \(\pounds 2\) for each kilometre. What is the cost of a 7 km journey?
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B: \(\pounds 17\)
\(3 + 2 \times 7 = 17\).
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6
On a graph of taxi cost against distance, what does the \(y\)-intercept mean?
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A: The fixed starting charge
The intercept is the cost for 0 km.
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7
A container gets wider towards the top and is filled at a steady rate. What happens to the depth-time graph?
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D: It rises more and more slowly, so it flattens
The wider the container, the more slowly the depth rises.
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8
A velocity-time graph is a triangle that rises from 0 to 8 m/s in 5 seconds. What distance does it show?
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C: 20 m
Area \(= \dfrac{1}{2} \times 5 \times 8 = 20\).
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9
How can you estimate the speed at one moment from a curved distance-time graph?
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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.