Printable · GCSE Higher · ages 14-16
Real-life graphs and kinematics worksheet — GCSE Higher
Fifteen questions on "real-life graphs and kinematics" — DfE statement A14. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
Answer key: Real-life graphs and kinematics worksheet — GCSE Higher
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- (a) 45 — The ratio between the two given points, one power of x apart, gives b: 405 ÷ 135 = 3, so b = 3. Substituting back, at x = 1, y = A × b, so 135 = A × 3, giving A = 45. Giving the common ratio b itself instead of A confuses which unknown was asked for and produces 3 — wrong, because the question asks for A, not b. Giving 135 instead treats the first given point as the y-intercept and reads A off it directly — wrong, because that point is at x = 1, not x = 0, so 135 is A × b, not A. Assuming A equals b⁰ = 1 by itself, rather than substituting a known point to solve for A, gives 1 — wrong, because b⁰ is always 1 regardless of A; A must be found using an actual (x, y) pair from the graph.
- (a) 2 m/s² — Method: acceleration = change in speed ÷ time = (17 − 5) ÷ 6 = 12 ÷ 6 = 2 m/s². Distractor origins: 12 m/s² stops after finding the change in speed and forgets to divide by the time; 22 m/s² adds the two speeds instead of subtracting them, and also forgets to divide by time (5 + 17 = 22); 72 m/s² multiplies the change in speed by the time instead of dividing (12 × 6 = 72).
- (a) It gets closer to zero but never actually reaches it. — y = 60 × (0.5)ˣ is always positive, however large x is, because a positive number raised to any power stays positive. As x increases, (0.5)ˣ gets smaller and smaller but never equals zero, so y approaches zero without ever reaching it. Believing the curve reaches zero when x = 60, because 60 is the starting value, mixes up the y-intercept with a stopping point — an exponential decay curve like this one never actually reaches the x-axis, so this is wrong. Believing the curve goes negative for large x is wrong because multiplying a positive number by (0.5) any number of times can never produce a negative result. Believing the curve levels off at y = 0.5, confusing the base of the exponential with its eventual level, is wrong: the base only controls how fast the curve falls, not where it settles — this curve settles at y = 0, not y = 0.5. Whenever you sketch an exponential decay curve, draw it approaching the x-axis but never touching it.
- (d) 400 m — The distance travelled equals the area under the speed-time graph. The first 10 seconds form a triangle with base 10 and height 20, giving an area of 0.5 × 10 × 20 = 100 m. The next 15 seconds form a rectangle with base 15 and height 20, giving an area of 15 × 20 = 300 m. The total distance is 100 + 300 = 400 m. The option 500 m treats the whole 25 seconds as travelled at the constant 20 m/s, ignoring that the speed was building up during the first 10 seconds: 25 × 20 = 500. The option 300 m only counts the constant-speed section and forgets the triangle section entirely. The option 200 m comes from working out the triangle's area without halving it (10 × 20 = 200) and forgetting the rectangle altogether.
- (d) 20 km/h — Method: total distance = 9 + 21 = 30 km. Total time = 15 + 45 + 30 = 90 minutes = 1.5 hours. Average speed = total distance ÷ total time = 30 ÷ 1.5 = 20 km/h. Distractor origins: 40 km/h ignores the 45-minute stop, dividing the total distance by the travel time only (30 km ÷ 45 minutes = 40 km/h); 39 km/h averages the two section speeds, 36 km/h and 42 km/h, instead of using total distance over total time; 36 km/h uses only the first section's speed (9 km in 15 minutes), ignoring the second section and the stop.
- (c) £8,000 — The multiplier for one year is 1 − 0.18 = 0.82, so 0.82 × 0.82 = 0.6724. Original value = 5,379.20 ÷ 0.6724 = £8,000. Treating the total drop as simple depreciation (18% × 2 = 36% lost in total, so dividing by 0.64 instead of 0.6724) gives 5,379.20 ÷ 0.64 = £8,405 (nearest pound) — wrong, because the depreciation compounds year by year rather than adding up. Dividing by 0.82 only once, forgetting the second year entirely, gives 5,379.20 ÷ 0.82 = £6,560 — wrong, because two years of depreciation have passed, not one. Multiplying by 0.6724 instead of dividing by it — repeating the depreciation forward instead of reversing it — gives 5,379.20 × 0.6724 = £3,617 (nearest pound) — wrong, because to find the ORIGINAL value you must undo the depreciation, not apply it again.
- (a) 3.5 km/h — Average speed is total distance ÷ total time. Total distance = 5 + 9 = 14 km. Total time, including the rest, = 1 + 1 + 2 = 4 hours. So average speed = 14 ÷ 4 = 3.5 km/h. Leaving out the 1 hour rest and dividing by only the 3 hours of walking gives 14 ÷ 3 = 4.67 km/h. Averaging the two separate speeds, 5 km/h and 4.5 km/h, instead of using total distance over total time, gives (5 + 4.5) ÷ 2 = 4.75 km/h. Dividing only the second leg's distance by the total time, 9 ÷ 4 = 2.25 km/h, ignores the first leg's distance.
- (a) 36 km/h — Speed is distance ÷ time. Convert 25 minutes to hours: 25 ÷ 60 = 5/12 hours. Then 15 ÷ (5/12) = 15 × 12/5 = 36 km/h. Dividing 15 by 25 without converting minutes to hours gives 0.6 km/h. Multiplying 15 by 25/60 instead of dividing by it gives 6.25 km/h. Working out the speed in km per minute, 0.6, and then multiplying by 24 instead of 60 gives 14.4 km/h.
- (c) 6 km/h — Total distance = 8 + 4 = 12 km. Total time = 1 hour + 0.5 hours resting + 0.5 hours = 2 hours. Average speed = total distance ÷ total time = 12 ÷ 2 = 6 km/h. A speed of 8 km/h comes from leaving the resting time out of the total time: 12 ÷ 1.5 = 8. A speed of 4 km/h comes from dividing by too much time, such as double-counting the rest period: 12 ÷ 3 = 4. A speed of 12 km/h simply gives the total distance and forgets to divide by the total time at all.
- (a) 12 km/h — Method: convert 15 minutes to hours: 15 ÷ 60 = 0.25 h. Speed = distance ÷ time = 3 ÷ 0.25 = 12 km/h. Distractor origins: 9 km/h is the speed for the second section (3 km in 20 minutes) instead of the first; 8 km/h is the average speed for the whole journey (6 km in 45 minutes) instead of just the first section; 0.2 km/h divides 3 km by 15 without converting the minutes into hours.
- (b) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (b) 5 cm³ — When P = 32, V = 480 ÷ 32 = 15 cm³. When P = 24, V = 480 ÷ 24 = 20 cm³. The increase in volume is 20 − 15 = 5 cm³. Giving only the second volume, 20 cm³, forgets to subtract the first volume. Giving only the first volume, 15 cm³, answers the wrong part of the question. Subtracting the two pressures instead of the two volumes, 32 − 24 = 8, mixes up which quantity the question asks for.
- (a) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
- (b) 58.8 km/h — Total distance = 100 + 47 = 147 km. Total time = 1.5 + 0.25 + 0.75 = 2.5 hours. Average speed = 147 ÷ 2.5 = 58.8 km/h. Leaving out the 0.25 hours of waiting from the total time gives 147 ÷ 2.25 = 65.3 km/h (1 d.p.) — wrong, because the train is stationary but time is still passing on the whole journey. Using only the first leg gives 100 ÷ 1.5 = 66.7 km/h (1 d.p.) — wrong, because it ignores the second leg of the journey entirely. Working out each leg's own speed (100 ÷ 1.5 = 66.7 km/h and 47 ÷ 0.75 = 62.7 km/h) and then averaging those two speeds gives 64.7 km/h (1 d.p.) — wrong, because the average of two speeds over DIFFERENT times is not the same as total distance divided by total time.
- (c) 72 m — The first section is a triangle with base 4 s and height 8 m/s, area = 1/2 × 4 × 8 = 16 m. The middle section is a rectangle with base 6 s and height 8 m/s, area = 6 × 8 = 48 m. The final section is a triangle with base 2 s and height 8 m/s, area = 1/2 × 2 × 8 = 8 m. Total distance = 16 + 48 + 8 = 72 m. Treating both sloping sections as full rectangles instead of triangles — forgetting the factor of a half — gives 4 × 8 + 6 × 8 + 2 × 8 = 96 m, which is wrong because a sloping speed-time section is a triangle, not a rectangle. Leaving out the final, decreasing section entirely gives 16 + 48 = 64 m, which is wrong because the cyclist is still moving and covering distance while slowing down. Leaving out the middle, constant-speed section gives 16 + 8 = 24 m, which is wrong because the cyclist covers the most distance of all during the 6 seconds of constant speed.
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