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.
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Answer key: Gradients and areas under graphs worksheet — GCSE Higher
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- (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.
- (a) Underestimate: the chord lies below the curve — For a curve that is concave down (bending downward), the straight line joining any two points on the curve lies BELOW the curve. The trapezium's top edge is this straight line, so the trapezium's area is smaller than the true area under the curve — the trapezium rule UNDERESTIMATES the distance in this case. This is because a curve whose gradient keeps decreasing bends away from any chord joining two of its points, dropping the chord below every point of the curve in between. The reverse, a chord lying above the curve, is the rule for a concave-up curve, not this one — check which way the curve bends before deciding. The trapezium rule does not always overestimate or underestimate, and it is not exact except when the graph really is a straight line.
- (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.
- (a) Overestimates — trapezium edges lie above the curve. — For a concave-up curve, each chord connecting two points on the curve lies ABOVE the curve between those points, because the curve bends away from the chord underneath it. The trapezium rule uses these chords as the top edges of each trapezium, so the trapeziums cover a larger region than the true area under the curve — the estimate is an overestimate. Saying it underestimates because the edges lie below the curve gets the direction backwards: for a concave-up curve the edges lie above, not below. Saying it overestimates 'because more strips always increase it' gives the right verdict for the wrong reason — using MORE, narrower strips makes a trapezium estimate more accurate and brings it closer to the true area, it does not simply increase the total. Saying the estimate is exact ignores that straight edges cannot follow a curved line perfectly; there is always a gap between the chord and the curve on a bend, which is exactly what makes the rule an estimate rather than an exact calculation.
- (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.
- (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.
- (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) 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.
- (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.
- (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) 10 m/s — A symmetric chord gradient uses the two points either side of t = 3: (2, 40) and (4, 60). The gradient is the change in height divided by the change in time: (60 − 40) ÷ (4 − 2) = 20 ÷ 2 = 10 m/s. Using only the values either side of one gap, (2, 40) and (3, 54), instead of the full symmetric chord, gives (54 − 40) ÷ (3 − 2) = 14 m/s. Getting the correct number but dropping the time unit, leaving only metres, gives 10 m. Dividing time by height instead of height by time inverts the calculation to (4 − 2) ÷ (60 − 40) = 0.1 s/m.
- (b) 14 m/s — Average speed for a whole journey is total distance divided by total time: 350 ÷ 25 = 14 m/s. Multiplying the distance and time instead of dividing gives 350 × 25 = 8750 m/s. Adding the distance and time instead of dividing gives 350 + 25 = 375 m/s. Inverting the division, working out time divided by distance, gives 25 ÷ 350, which rounds to 0.07 m/s.
- (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.
- (c) Fastest at t = 2 min — steepest gradient. — The rate of cooling is given by the size (magnitude) of the gradient, ignoring its sign — the steeper the tangent, the faster the temperature is changing. Of −8, −3 and −0.5, the gradient −8 has the greatest magnitude, so the tea is cooling fastest at t = 2 minutes. 'Fastest at t = 20 min — largest gradient' confuses the signed value with the size of the rate: −0.5 is the largest NUMBER of the three, but it's the smallest in magnitude, meaning the tea is barely cooling at all by then. 'Cools at the same rate throughout' ignores that the three gradients are different sizes, not just all negative. 'Fastest at t = 10 min — the middle reading' isn't a mathematical reason at all — the gradients themselves have to be compared, not their position in the list.
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