Printable · GCSE Higher · ages 14-16
Gradients and areas under graphs worksheet — GCSE Higher
Fifteen questions on "gradients and areas under graphs" — DfE statement A15. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Higher only
Gradients and areas under graphs worksheet — GCSE Higher
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- 1.The table shows the speed, in m/s, of a car at times, in seconds, all 2 seconds apart: 0 at t = 0, 3 at t = 2, 8 at t = 4, 15 at t = 6, 24 at t = 8. A student uses the trapezium rule with these five readings to estimate the distance the car travels. By finding the second differences of the speed values, decide whether this trapezium estimate is an overestimate or an underestimate of the true distance.
- 2.The curve y = x² − 3x. Use the chord between x = 1 and x = 4 to estimate the gradient of the curve at x = 2.5.y = x² − 3x
- 3.On a distance-time graph, a horizontal line segment shows a period when the graph's gradient is 0. What does this tell you about the journey during that time?
- 4.The tangent to a curve at the point where x = 4 passes through the points (2, 5) and (6, 21). Use these two points to estimate the gradient of the curve at x = 4.
- 5.A cyclist's velocity increases from 3 m/s to 11 m/s over 5 seconds, at a constant rate. Work out the cyclist's acceleration, in m/s².
- 6.A car is decelerating. The tangent to its velocity-time graph at t = 12 seconds passes through the points (8, 22) and (16, 6), where velocity is in m/s and time is in seconds. Work out the gradient of this tangent, in m/s².
- 7.A tap fills a tank at a varying rate. A graph shows the rate of flow, in litres per minute, against time, in minutes. What does the area under this graph represent?
- 8.The table shows the velocity, in m/s, of a train at various times, in seconds: 0 at t = 0, 20 at t = 4, 35 at t = 10, 27 at t = 14. Assuming the velocity changes at a constant rate between each pair of readings, estimate the distance the train travels between t = 0 and t = 12, using the trapezium rule.
- 9.A car's speed, in m/s, during a journey is: 0 at t = 0 s, 8 at t = 4 s, and 8 (constant) from t = 4 s to t = 10 s, increasing at a constant rate between t = 0 and t = 4. Estimate the total distance the car travels, using the areas of a triangle and a rectangle.
- 10.A cyclist accelerates uniformly from rest to 6 m/s in 4 seconds, travels at a constant 6 m/s for 10 seconds, then decelerates uniformly to rest in 3 seconds. Work out the total distance the cyclist travels.
- 11.The area under a speed-time graph between t = 0 and t = 6 seconds is estimated using three strips of equal width, using the speeds, in m/s, at t = 0, 2, 4 and 6: 0, 5, 9 and 12. Using the trapezium rule with these three trapezia, estimate the distance travelled.
- 12.A speed-time graph shows a constant speed of 15 m/s for 20 seconds. Work out the distance travelled, using the area under the graph.
- 13.The tangent to a curve at the point (5, 2) is the line y = mx + c. This tangent crosses the y-axis at (0, −8). Work out the gradient, m, of the tangent.
- 14.The table shows the rate at which water flows from a tank, in litres per minute, over a 6-minute period: 12 at t = 0 min, 12 at t = 2 min, 3 at t = 5 min, and 0 at t = 6 min. Between t = 0 and t = 2 the rate is constant. Between t = 2 and t = 5, and between t = 5 and t = 6, the rate decreases at a constant rate in each section. Estimate the total volume of water that has flowed out, using the areas of a rectangle and two trapeziums.
- 15.A tap's flow rate, in litres per minute, is plotted against time, in minutes, on a graph. What are the units of the area under this graph?
Answer key
- (d) Overestimate — the curve bends upward (convex). — The first differences of the speeds are 3, 5, 7 and 9, so the second differences are 2, 2 and 2 — constant and positive, which means the speed-time graph curves upwards (is convex). On a convex curve, each straight chord used by the trapezium rule lies above the curve, so the trapezium rule overestimates the true distance. 'Underestimate — the curve bends upward' states the same correct geometry but gets the conclusion backwards — a chord above the curve means too much area is counted, not too little. 'Overestimate — the speed values are increasing' uses the wrong evidence: increasing speed alone doesn't tell you whether the curve bends up or down, only the second differences do. 'Underestimate — second differences are constant' confuses a constant second difference with a steady rate of change in speed, which isn't what the second difference of a speed-time table measures.
- (c) 2 — At x = 1, y = 1² − 3(1) = 1 − 3 = −2. At x = 4, y = 4² − 3(4) = 16 − 12 = 4. Gradient of the chord = change in y ÷ change in x = (4 − (−2)) ÷ (4 − 1) = 6 ÷ 3 = 2. Writing down the change in y, 6, and stopping there without dividing by the change in x gives 6. Losing the negative sign on y = −2 at x = 1 and treating it as +2 gives (4 − 2) ÷ (4 − 1) = 2 ÷ 3 = 2/3. Dividing the wrong way round, change in x ÷ change in y, gives (4 − 1) ÷ (4 − (−2)) = 3 ÷ 6 = 1/2.
- (c) The object was stationary (not moving) — On a distance-time graph, the gradient at any point represents the speed at that point. A gradient of 0 means distance is not changing over time, so the object is stationary. A straight, sloped line (not flat) shows constant nonzero speed; a curve bending one way shows acceleration or deceleration; a flat section is not a maximum speed — it is no speed at all. Read the shape of the graph, not just how steep it looks.
- (a) 4 — The gradient of a straight line through two points is the change in y divided by the change in x. Change in y = 21 − 5 = 16. Change in x = 6 − 2 = 4. Gradient = 16 ÷ 4 = 4. Dividing x by y instead of y by x gives 0.25; forgetting to divide by the change in x at all leaves 16; dividing by only one of the two x-coordinates, 16 ÷ 2 = 8, uses the wrong denominator.
- (b) 1.6 m/s² — Acceleration is the change in velocity divided by the time taken: (11 − 3) ÷ 5 = 8 ÷ 5 = 1.6 m/s². Forgetting to subtract the initial velocity and dividing the final velocity by the time instead gives 11 ÷ 5 = 2.2 m/s². Inverting the fraction, dividing the time by the change in velocity, gives 5 ÷ 8 = 0.625 m/s². Finding the change in velocity, 8 m/s, but stopping without dividing by the time gives 8 m/s².
- (a) −2 m/s² — The gradient of a line through two points is the change in the y-value divided by the change in the x-value: (6 − 22)/(16 − 8) = −16/8 = −2 m/s². The negative sign confirms the deceleration mentioned in the question. Writing 2 m/s² comes from subtracting the velocities in the wrong order, using (22 − 6) instead of (6 − 22), which loses the sign that shows the car is slowing down. Writing −0.5 m/s² comes from inverting the fraction, dividing the change in time by the change in velocity instead of the other way round. Writing −16 m/s² comes from finding the change in velocity, −16, but forgetting to divide by the change in time, 8 seconds.
- (b) The total volume of water, in litres, that has flowed in. — On a rate-time graph, the y-axis is in litres per minute and the x-axis is in minutes; multiplying a rate by a time gives litres per minute × minutes = litres, a total volume. So the area under the graph represents the total volume of water that has flowed in. Thinking the area itself represents the rate, rather than what the rate accumulates to, gives the wrong claim about the average rate of flow. Confusing the area with the gradient of the graph — which measures how the rate is changing — gives the wrong claim about litres per minute squared. Ignoring the flow-rate axis and focusing only on the time axis gives the wrong claim that the area is simply the total time.
- (a) 271 m — The reading at t = 12 isn't in the table, so it must be interpolated between t = 10 (35 m/s) and t = 14 (27 m/s): the velocity falls by 8 m/s over the 4-second gap, so over 2 seconds it falls by 4 m/s, giving 35 − 4 = 31 m/s. The trapezium rule then uses three strips of unequal width, 4, 6 and 2 seconds, with areas of 40, 165 and 66: 40 + 165 + 66 = 271 m. Writing 267 m comes from skipping the interpolation and using the table's t = 14 reading, 27, directly as the height at t = 12. Writing 282 m comes from treating all three strips as though they were 4 seconds wide, instead of using the true gaps of 4, 6 and 2 seconds between the readings. Writing 275 m comes from using the t = 10 reading, 35, at both ends of the final strip, instead of interpolating a new value at t = 12.
- (c) 64 m — The distance travelled is the area under the speed-time graph. From t = 0 to t = 4, the shape is a triangle with base 4 and height 8, area 1/2 × 4 × 8 = 16. From t = 4 to t = 10, the shape is a rectangle with base 6 and height 8, area 6 × 8 = 48. Total distance: 16 + 48 = 64 m. Treating the whole 10 seconds as a single trapezium with parallel sides 0 and 8 and width 10, instead of splitting it into the triangle and rectangle, gives 40 m. Ignoring the acceleration phase completely and assuming the car travels at a constant 8 m/s for all 10 seconds gives 8 × 10 = 80 m. Misreading the second interval as running from t = 4 to t = 8 instead of t = 4 to t = 10 gives a rectangle area of 4 × 8 = 32, plus the correct triangle of 16, totalling 48 m.
- (c) 81 m — Split the velocity-time graph into its three phases. The accelerating phase (0 to 4 s) is a triangle: 1/2 × 4 × 6 = 12. The constant phase (4 s at 6 m/s, for 10 s) is a rectangle: 10 × 6 = 60. The decelerating phase (3 s) is a triangle: 1/2 × 3 × 6 = 9. Total distance: 12 + 60 + 9 = 81 m. Forgetting the final decelerating phase entirely and adding only the first two areas gives 12 + 60 = 72 m. Treating all three phases as rectangles, forgetting to halve either triangular phase, gives 4 × 6 + 10 × 6 + 3 × 6 = 102 m. Halving only the deceleration triangle correctly but treating the acceleration phase as a rectangle too gives 4 × 6 + 60 + 9 = 93 m.
- (d) 40 m — Each trapezium has width 2. Its area is width × the average of its two heights. Strip 1: average of 0 and 5 is 2.5, so area = 2 × 2.5 = 5. Strip 2: average of 5 and 9 is 7, so area = 2 × 7 = 14. Strip 3: average of 9 and 12 is 10.5, so area = 2 × 10.5 = 21. Total distance = 5 + 14 + 21 = 40 m. Leaving out the division by 2 in the averaging step doubles every strip, giving 80 m instead of 40 m. Using only the LEFT height of each strip as a rectangle, 0 × 2 + 5 × 2 + 9 × 2 = 28, or only the RIGHT height, 5 × 2 + 9 × 2 + 12 × 2 = 52, both ignore that the graph curves between the two ends of each strip — always average the two heights of a trapezium, never use just one of them.
- (d) 300 m — For a constant speed, the speed-time graph is a horizontal line, and the area underneath is a rectangle: distance = speed × time = 15 × 20 = 300 m. Adding the two numbers instead of multiplying gives 35 m; dividing instead of multiplying gives 1.33 m; halving the product, as you would for a triangle, gives 7.5 m — but this section of the graph is a rectangle, not a triangle, so there is no halving to do.
- (d) 2 — The gradient between two points on a line is the change in y divided by the change in x. Using (5, 2) and (0, −8): (2 − (−8)) ÷ (5 − 0) = 10 ÷ 5 = 2. Inverting the fraction, dividing the change in x by the change in y instead, gives 5 ÷ 10 = 0.5. Reversing the order of the x-values in the denominator, giving (2 − (−8)) ÷ (0 − 5) = 10 ÷ (−5), gives −2. Misreading the y-intercept as 8 instead of −8, giving (2 − 8) ÷ 5 = −6 ÷ 5, gives −1.2.
- (a) 48 litres — Split the area into three sections. The rectangle (t = 0 to 2) has area 2 × 12 = 24. The first trapezium (t = 2 to 5, parallel sides 12 and 3, width 3) has area 1/2 × (12 + 3) × 3 = 22.5. The second trapezium (t = 5 to 6, parallel sides 3 and 0, width 1) has area 1/2 × (3 + 0) × 1 = 1.5. Total volume: 24 + 22.5 + 1.5 = 48 litres. Forgetting the rectangle and adding only the two trapeziums gives 22.5 + 1.5 = 24 litres. Using a single trapezium across the whole 6 minutes, with parallel sides 12 and 0, ignoring that the first 2 minutes are constant, gives 1/2 × (12 + 0) × 6 = 36 litres. Correctly finding the rectangle and the first trapezium but forgetting to halve the second trapezium, using (3 + 0) × 1 = 3 instead of 1.5, gives 24 + 22.5 + 3 = 49.5 litres.
- (a) Litres — The area under a graph is found by multiplying a y-value by an x-value, so its units are the y-axis units multiplied by the x-axis units: litres per minute × minutes = litres, since the 'per minute' cancels with the 'minutes'. Answering 'litres per minute' keeps the y-axis units unchanged, as if multiplying by time did nothing to the units at all. Answering 'litres per minute squared' treats the x-axis as also being measured 'per minute', squaring a unit that should instead cancel. Answering 'minutes' keeps only the x-axis units and drops the rate altogether.
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