Exam questions · Maths
Graphs
- 30 exam questions
- 93 marks
- 45 quick checks
Straight-Line Graphs
Just this lesson-
1 Write down [2 marks]
Write down the equation of the line with gradient 3 that crosses the \(y\)-axis at \((0, -4)\).
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Model answer
Using \(y = mx + c\) with \(m = 3\) and \(c = -4\), the equation is \(y = 3x - 4\).
Mark scheme
- \(y = 3x + c\) or \(y = mx - 4\) — B1
- \(y = 3x - 4\) — B1
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2 Complete [3 marks]
(a) Complete the table of values for \(y = 2x + 3\). \(x = -2, -1, 0, 1, 2\) [2 marks] (b) Does the point \((6, 15)\) lie on the line \(y = 2x + 3\)? Show how you know. [1 mark]
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Model answer
(a) The values are \(-1, 1, 3, 5, 7\). (b) When \(x = 6\), \(y = 2 \times 6 + 3 = 15\), so the point lies on the line.
Mark scheme
- (a) At least three correct values — M1
- (a) \(-1, 1, 3, 5, 7\) — A1
- (b) Yes, with \(2 \times 6 + 3 = 15\) — B1
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3 Work out [4 marks]
The diagram shows a straight line through the points \(P\) and \(Q\). (a) Work out the gradient of the line. [2 marks] (b) Write down the equation of the line. [2 marks]
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Model answer
(a) \(\dfrac{1 - 3}{6 - 2} = \dfrac{-2}{4} = -\dfrac{1}{2}\). (b) The line crosses the \(y\)-axis at 4, so \(y = -\dfrac{1}{2}x + 4\).
Mark scheme
- (a) \(\dfrac{1 - 3}{6 - 2}\) — M1
- (a) \(-\dfrac{1}{2}\) — A1
- (b) \(y = -\dfrac{1}{2}x + c\) or \(y = mx + 4\) — M1
- (b) \(y = -\dfrac{1}{2}x + 4\) — A1
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4 Show that [3 marks]
Line \(A\) has equation \(y = 2x - 1\). Line \(B\) passes through \((0, 5)\) and \((2, 9)\). Show that lines \(A\) and \(B\) are parallel.
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Model answer
The gradient of line \(B\) is \(\dfrac{9 - 5}{2 - 0} = 2\). The gradient of line \(A\) is also 2. Parallel lines have the same gradient, so the lines are parallel.
Mark scheme
- \(\dfrac{9 - 5}{2 - 0}\) — M1
- Gradient of \(B\) is 2 — A1
- States that the gradients are equal so the lines are parallel — Q1
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5 Write down [2 marks]
The lines \(x = 3\) and \(y = -2\) cross at a point. (a) Write down the coordinates of the point. [1 mark] (b) Which of the two lines is vertical? [1 mark]
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Model answer
(a) Where \(x = 3\) and \(y = -2\), the point is \((3, -2)\). (b) The line \(x = 3\) is vertical.
Mark scheme
- (a) \((3, -2)\) — B1
- (b) \(x = 3\) — B1
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6 Work out [3 marks]
A line has equation \(2x + y = 10\). (a) Work out its gradient. [2 marks] (b) Write down the coordinates of the point where it crosses the \(x\)-axis. [1 mark]
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Model answer
(a) \(y = -2x + 10\), so the gradient is \(-2\). (b) When \(y = 0\), \(2x = 10\) and \(x = 5\), so the point is \((5, 0)\).
Mark scheme
- (a) \(y = -2x + 10\) — M1
- (a) \(-2\) — A1
- (b) \((5, 0)\) — B1
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 Work out [2 marks]
Work out the coordinates of the midpoint of \((-4, 3)\) and \((6, 9)\).
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Model answer
\(\left(\dfrac{-4 + 6}{2}, \dfrac{3 + 9}{2}\right) = (1, 6)\).
Mark scheme
- One coordinate correct — M1
- \((1, 6)\) — A1
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2 Work out [6 marks]
The diagram shows a straight line through the points \(A\) and \(B\). (a) Work out the gradient of \(AB\). [2 marks] (b) Find the equation of \(AB\). [2 marks] (c) Work out the coordinates of the midpoint of \(AB\). [2 marks]
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Model answer
(a) \(\dfrac{-1 - 5}{4 - (-2)} = \dfrac{-6}{6} = -1\). (b) \(y = -x + c\) with \((4, -1)\) gives \(-1 = -4 + c\), so \(c = 3\) and \(y = -x + 3\). (c) \(\left(\dfrac{-2 + 4}{2}, \dfrac{5 + (-1)}{2}\right) = (1, 2)\).
Mark scheme
- (a) \(\dfrac{-1 - 5}{4 - (-2)}\) — M1
- (a) \(-1\) — A1
- (b) \(y = -x + c\) with a point substituted — M1
- (b) \(y = -x + 3\) — A1
- (c) One coordinate correct — M1
- (c) \((1, 2)\) — A1
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3 Find [3 marks]
A line has gradient \(-2\) and passes through the point \((3, 1)\). Find the equation of the line.
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Model answer
\(y = -2x + c\). Putting in \((3, 1)\) gives \(1 = -6 + c\), so \(c = 7\) and \(y = -2x + 7\).
Mark scheme
- \(y = -2x + c\) — M1
- \(1 = -2 \times 3 + c\) — M1
- \(y = -2x + 7\) — A1
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4 Show that [2 marks]
Show that the lines \(2y = 4x + 5\) and \(y = 2x - 3\) are parallel.
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Model answer
Dividing the first equation by 2 gives \(y = 2x + 2.5\), which has gradient 2. The second line also has gradient 2, so they are parallel.
Mark scheme
- \(y = 2x + 2.5\) or gradient 2 found — M1
- Both gradients are 2, so they are parallel — Q1
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5 Find [3 marks]
The line \(P\) has equation \(y = \dfrac{1}{2}x + 1\). The line \(Q\) is perpendicular to \(P\) and passes through \((0, 4)\). Find an equation of \(Q\).
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Model answer
The gradient of \(Q\) is \(-2\), because \(\dfrac{1}{2} \times (-2) = -1\). It crosses the \(y\)-axis at 4, so \(y = -2x + 4\).
Mark scheme
- Gradient \(-2\) seen — M1
- \(y = -2x + c\) or \(c = 4\) seen — M1
- \(y = -2x + 4\) — A1
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6 Work out [4 marks]
\(A\) is the point \((-1, 2)\) and \(B\) is the point \((5, 10)\). (a) Work out the length of \(AB\). [3 marks] (b) Work out the midpoint of \(AB\). [1 mark]
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Model answer
(a) The differences are 6 and 8, so \(AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\). (b) The midpoint is \(\left(\dfrac{-1 + 5}{2}, \dfrac{2 + 10}{2}\right) = (2, 6)\).
Mark scheme
- (a) Differences 6 and 8 — M1
- (a) \(\sqrt{6^2 + 8^2}\) — M1
- (a) 10 — A1
- (b) \((2, 6)\) — B1
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 \(y = x^2 + 1\). \(x = -2, -1, 0, 1, 2\)
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Model answer
The values are \(5, 2, 1, 2, 5\).
Mark scheme
- At least three correct values — M1
- \(5, 2, 1, 2, 5\) — A1
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2 Write down [4 marks]
The diagram shows the graph of \(y = x^2 - 6x + 5\). (a) Use the graph to solve \(x^2 - 6x + 5 = 0\). [2 marks] (b) Write down the coordinates of the turning point. [1 mark] (c) Write down the equation of the line of symmetry. [1 mark]
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Model answer
(a) The curve crosses the \(x\)-axis at 1 and 5, so \(x = 1\) and \(x = 5\). (b) The lowest point is \((3, -4)\). (c) The line of symmetry is \(x = 3\).
Mark scheme
- (a) One of \(x = 1\) or \(x = 5\) — M1
- (a) \(x = 1\) and \(x = 5\) — A1
- (b) \((3, -4)\) — B1
- (c) \(x = 3\) — B1
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3 Work out [3 marks]
A curve has equation \(y = x^2 - 2x - 8\). (a) Write down the \(y\)-intercept. [1 mark] (b) Work out the coordinates of the points where the curve crosses the \(x\)-axis. [2 marks]
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Model answer
(a) Put \(x = 0\): \(y = -8\). (b) \(x^2 - 2x - 8 = (x - 4)(x + 2) = 0\), so the points are \((4, 0)\) and \((-2, 0)\).
Mark scheme
- (a) \(-8\) — B1
- (b) \((x - 4)(x + 2)\) — M1
- (b) \((4, 0)\) and \((-2, 0)\) — A1
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4 Work out [3 marks]
The graph of \(y = x^2 - 4x\) crosses the \(x\)-axis at \(x = 0\) and \(x = 4\). Work out the coordinates of its turning point.
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Model answer
The turning point is halfway between the roots, at \(x = 2\). Then \(y = 2^2 - 4 \times 2 = -4\). The turning point is \((2, -4)\).
Mark scheme
- \(x = 2\) seen — M1
- \(2^2 - 4 \times 2\) — M1
- \((2, -4)\) — A1
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5 Work out [4 marks]
(a) Write \(x^2 + 6x + 1\) in the form \((x + a)^2 + b\). [2 marks] (b) Write down the coordinates of the turning point of the graph of \(y = x^2 + 6x + 1\). [1 mark] (c) Write down the minimum value of \(x^2 + 6x + 1\). [1 mark]
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Model answer
(a) \((x + 3)^2 - 9 + 1 = (x + 3)^2 - 8\). (b) \((-3, -8)\). (c) \(-8\).
Mark scheme
- (a) \((x + 3)^2\) seen — M1
- (a) \((x + 3)^2 - 8\) — A1
- (b) \((-3, -8)\) — B1
- (c) \(-8\) — B1
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6 Show that [3 marks]
A parabola crosses the \(x\)-axis at \(x = -3\) and \(x = 1\). (a) Write down the equation of its line of symmetry. [1 mark] The equation of the parabola is \(y = x^2 + 2x - 3\). (b) Show that the turning point is \((-1, -4)\). [2 marks]
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Model answer
(a) \(x = -1\), halfway between \(-3\) and 1. (b) When \(x = -1\), \(y = (-1)^2 + 2 \times (-1) - 3 = 1 - 2 - 3 = -4\), so the turning point is \((-1, -4)\).
Mark scheme
- (a) \(x = -1\) — B1
- (b) \((-1)^2 + 2 \times (-1) - 3\) — M1
- (b) \(-4\) with a conclusion — Q1
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 \(y = 2^x\), \(y = 4 - x^2\) and \(y = 3 - x\). Match each equation to the correct graph. [3 marks]
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Model answer
Graph \(A\) is an upside-down U, so \(y = 4 - x^2\). Graph \(B\) rises quickly and stays above the \(x\)-axis, so \(y = 2^x\). Graph \(C\) is a falling straight line, so \(y = 3 - x\).
Mark scheme
- \(A\) is \(y = 4 - x^2\) — B1
- \(B\) is \(y = 2^x\) — B1
- \(C\) is \(y = 3 - x\) — B1
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2 Complete [2 marks]
Complete the table of values for \(y = x^3 + 1\). \(x = -2, -1, 0, 1, 2\)
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Model answer
The values are \(-7, 0, 1, 2, 9\).
Mark scheme
- At least three correct values — M1
- \(-7, 0, 1, 2, 9\) — A1
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3 Complete [3 marks]
(a) Complete the table of values for \(y = \dfrac{6}{x}\). \(x = 1, 2, 3, 6, -2, -3\) [2 marks] (b) How many separate parts, or branches, does the graph of \(y = \dfrac{6}{x}\) have? [1 mark]
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Model answer
(a) The values are \(6, 3, 2, 1, -3, -2\). (b) The graph has 2 branches.
Mark scheme
- (a) At least four correct values — M1
- (a) \(6, 3, 2, 1, -3, -2\) — A1
- (b) 2 — B1
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4 Complete [3 marks]
(a) Complete the table of values for \(y = 3^x\). \(x = 0, 1, 2, 3\) [2 marks] (b) Write down the \(y\)-intercept of the graph of \(y = 3^x\). [1 mark]
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Model answer
(a) The values are \(1, 3, 9, 27\). (b) The \(y\)-intercept is 1, because \(3^0 = 1\).
Mark scheme
- (a) At least two correct values — M1
- (a) \(1, 3, 9, 27\) — A1
- (b) 1 — B1
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5 Work out [3 marks]
The graph of \(y = x^2 - 4\) is symmetrical. (a) Write down the equation of its line of symmetry. [1 mark] (b) Work out the coordinates of the points where the graph crosses the \(x\)-axis. [2 marks]
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Model answer
(a) The line of symmetry is the \(y\)-axis, \(x = 0\). (b) When \(y = 0\), \(x^2 = 4\), so \(x = 2\) or \(x = -2\). The points are \((2, 0)\) and \((-2, 0)\).
Mark scheme
- (a) \(x = 0\) — B1
- (b) \(x^2 = 4\) — M1
- (b) \((2, 0)\) and \((-2, 0)\) — A1
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6 Show that [3 marks]
A circle has equation \(x^2 + y^2 = 100\). (a) Write down the radius of the circle. [1 mark] (b) Show that the point \((6, 8)\) lies on the circle. [2 marks]
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Model answer
(a) The radius is \(\sqrt{100} = 10\). (b) \(6^2 + 8^2 = 36 + 64 = 100\), so the point lies on the circle.
Mark scheme
- (a) 10 — B1
- (b) \(6^2 + 8^2\) or \(36 + 64\) — M1
- (b) 100 with a conclusion — Q1
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 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]
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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
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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
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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
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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
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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
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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
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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.