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.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?
- 2.A tram sets off from a stop. Its velocity-time graph rises in a straight line from 0 m/s to 12 m/s over the first 8 seconds, then stays constant at 12 m/s for a further 10 seconds. Work out the total distance the tram travels in these 18 seconds, using the area under the graph.
- 3.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.
- 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 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.
- 6.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.
- 7.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.
- 8.A car accelerates uniformly from rest at 2.5 m/s² until it reaches a speed of 20 m/s, then travels at this constant speed for a further 30 seconds. Work out the total distance travelled.
- 9.A car's velocity–time graph is a straight line from (0 s, 4 m/s) rising to (6 s, V m/s), followed by a straight line falling from (6 s, V m/s) to (9 s, 0 m/s). The gradient of the second line is −8 m/s². Work out the total distance travelled between t = 0 and t = 9 seconds.
- 10.The table shows the speed, in m/s, of a cyclist at times, in seconds, 1 second apart: 0 at t = 0, 7 at t = 1, 12 at t = 2, 15 at t = 3, 16 at t = 4, 14 at t = 5, 9 at t = 6. Using the trapezium rule with strip width 1 second, estimate the distance the cyclist travels between t = 2 and t = 5 only.
- 11.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.
- 12.The tangent to a curve at the point (6, 1) has gradient −3. Work out the y-coordinate of the point where this tangent crosses the y-axis.
- 13.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.
- 14.A distance-time graph is a straight line from (0, 0) to (4, 100), where time is in hours and distance is in kilometres. Work out the gradient of the line.
- 15.A company's cost, in £, for producing x items is shown on a graph. The tangent to the curve at x = 50 passes through (30, 400) and (70, 800). Interpret the gradient of this tangent in the context of the company's costs.
Answer key
- (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.
- (d) 168 m — Split the area under the graph into two parts. The rising section (0 to 8 s) is a triangle: area = 0.5 × 8 × 12 = 48. The constant section, from 8 s to 18 s (10 s), is a rectangle: area = 12 × 10 = 120. Total distance = 48 + 120 = 168 m. Leaving out the 0.5 doubles the triangle, giving 8 × 12 + 120 = 216 m; ignoring the triangle section altogether gives only 120 m; averaging the start and final speeds over the whole 18 seconds, (0 + 12) ÷ 2 = 6, then 6 × 18 = 108 m, wrongly treats the tram as accelerating the whole time, when the graph is flat for the last 10 seconds.
- (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.
- (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.
- (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.
- (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) 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.
- (d) 680 m — First find how long the acceleration takes: acceleration = change in speed ÷ time, so 2.5 = 20 ÷ t, giving t = 20 ÷ 2.5 = 8 seconds. The distance during this phase is the area of a triangle with base 8 and height 20: 1/2 × 8 × 20 = 80 m. The distance during the constant-speed phase is 20 × 30 = 600 m, since distance = speed × time at a constant speed. The total distance is 80 + 600 = 680 m. Using the given 30 seconds for the acceleration phase as well, 1/2 × 30 × 20 = 300, plus the correct 600, gives 900 m — but 30 seconds is only stated for the constant-speed phase. Leaving out the 1/2 and using the full rectangle for the acceleration phase, 8 × 20 = 160, plus the correct 600, gives 760 m — the speed is not constant during acceleration, so this area is a triangle, not a rectangle. Swapping the two times round, and using 8 seconds for the constant-speed distance instead of 30, 20 × 8 = 160, plus the correct triangle area of 80, gives 240 m.
- (a) 120 m — The gradient of the second line is (0 − V)/(9 − 6) = −V/3, and this equals −8, so V = 24. The distance from t = 0 to t = 6 is the area of a trapezium with parallel sides 4 and 24 and width 6: 1/2 × (4 + 24) × 6 = 84. The distance from t = 6 to t = 9 is the area of a triangle with base 3 and height 24: 1/2 × 3 × 24 = 36. The total distance is 84 + 36 = 120 m. Using V = 8, treating the gradient's number as the missing velocity itself rather than solving −V/3 = −8 for V, gives a trapezium area of 1/2 × (4 + 8) × 6 = 36 and a triangle area of 1/2 × 3 × 8 = 12, a total of 48 m. Leaving out the 1/2 in the trapezium formula, (4 + 24) × 6 = 168, plus the correct triangle of 36, gives 204 m. Using the full 6 seconds as the triangle's base instead of the 3 seconds the second line actually lasts, 1/2 × 6 × 24 = 72, plus the correct trapezium of 84, gives 156 m.
- (d) 44 m — The trapezium rule between t = 2 and t = 5 uses only the speeds at t = 2, 3, 4 and 5 — 12, 15, 16 and 14 — with three strips of width 1: adding the first and last readings and twice the sum of the middle readings gives 12 + 14 + 2 × (15 + 16) = 88, and half of that is 44 m. Writing 68.5 m comes from applying the trapezium rule to the whole table, from t = 0 to t = 6, instead of restricting it to the interval t = 2 to t = 5 that the question asks for. Writing 28.5 m comes from forgetting to double the two middle readings, 15 and 16, in the trapezium rule formula. Writing 39 m comes from averaging only the first and last speeds in the interval, 12 and 14, and multiplying by the 3-second interval, ignoring the readings at t = 3 and t = 4 in between.
- (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.
- (b) 19 — A line through (x₁, y₁) with gradient m has equation y − y₁ = m(x − x₁). Substituting (6, 1) and m = −3: y − 1 = −3(x − 6), so y = −3x + 18 + 1, which simplifies to y = −3x + 19; at x = 0 this gives 19. Making a sign error when distributing, writing −3(x − 6) as −3x − 18 instead of −3x + 18, gives the wrong line y = −3x − 17, so −17 at x = 0. Forgetting to add the y₁ = 1 at the end, using y = −3(x − 6) alone, gives 18 at x = 0. Substituting the y-coordinate into the gradient term instead of using x, working out 1 + (−3 × 1), gives −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) 25 — Gradient = change in y ÷ change in x = (100 − 0) ÷ (4 − 0) = 100 ÷ 4 = 25. Dividing time by distance instead of distance by time gives 0.04; multiplying the two values instead of dividing gives 400; stopping at the change in distance, 100, forgets to divide by the change in time.
- (d) The cost increases by about £10 per extra item — Gradient = change in cost ÷ change in items = (800 − 400) ÷ (70 − 30) = 400 ÷ 40 = 10. The units of the gradient are £ per item, so the cost is increasing by about £10 for every extra item produced, near x = 50. Subtracting in the wrong order, (400 − 800) ÷ (70 − 30) = −10, gives the right size but the wrong sign — check which point comes first each time. Leaving out the division by 40 gives £400 per item; dividing the wrong way round, 40 ÷ 400 = 0.1, gives £0.10 per item.
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