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.
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Real-life graphs and kinematics worksheet — GCSE Higher
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- 1.A population of bacteria doubles every 3 hours. It starts at 800. Work out the population after 9 hours.
- 2.A train's distance-time graph shows it travelling 100 km at a steady speed in the first 1.5 hours, waiting at a signal for 0.25 hours, then travelling a further 47 km in 0.75 hours. Work out the average speed for the whole journey, correct to 1 decimal place.
- 3.A graph shows the total monthly cost of a mobile phone tariff against the number of minutes of calls. The line starts at £12 and stays level until 200 minutes, then rises by 5p for each further minute of calls. Use the graph to work out the total cost of a month in which 250 minutes of calls are made.
- 4.The graph of W = 840 ÷ L shows how the width, W metres, of a rectangular field of area 840 m² depends on its length, L metres. Use the relationship to work out the width when the length is 35 m, and hence the perimeter of the field.
- 5.A bacteria colony doubles every 4 hours and its size is recorded only every 4 hours. The recorded population 20 hours after the start is 9,600. After how many hours did the recorded population first exceed 1,000?
- 6.An exponential graph y = A × rˣ passes through the points (0, 8) and (2, 18). Work out the value of r, correct to 2 decimal places.
- 7.The mass of a chemical sample decays by 15% every hour. It starts at 500 g. Work out the mass remaining after 3 hours, correct to 1 decimal place.
- 8.A ferry's distance-time graph is a straight line from (0 minutes, 0 km) to (25 minutes, 15 km). Work out the ferry's speed in kilometres per hour.
- 9.A machine bought for £2,400 loses 12% of its value every year. After how many complete years does its value first fall below £1,500?
- 10.A cyclist's speed-time graph is described as follows: speed increases steadily from 0 m/s to 8 m/s over the first 4 seconds, stays constant at 8 m/s for the next 6 seconds, then decreases steadily to 0 m/s over the final 2 seconds. Work out the total distance travelled.
- 11.The graph of y = 60 × (0.5)ˣ is sketched for x ≥ 0. Which statement correctly describes what happens to the curve as x increases?
- 12.An exponential model has equation y = A × bˣ, where b > 1. Its graph passes through the points (1, 135) and (2, 405). Work out the value of A.
- 13.A hiker's distance-time graph shows the following: he walks 5 km in the first 1 hour at a steady speed, rests for 1 hour, then walks a further 9 km in the next 2 hours, also at a steady speed. Work out his average speed for the whole journey, in km/h.
- 14.The graph of P = 480 ÷ V shows how the pressure, P pascals, of a gas depends on its volume, V cm³, at constant temperature. Use the relationship to work out the volume when the pressure is 32 pascals, and then work out by how much the volume increases when the pressure drops to 24 pascals.
- 15.A cyclist's distance-time graph shows the following: she travels 12 km in the first 30 minutes at a steady speed, then rests for 30 minutes, then travels a further 8 km in the next 15 minutes. Work out her average speed, in km/h, for the whole journey.
Answer key
- (c) 6,400 — 9 hours contains 9 ÷ 3 = 3 whole periods of doubling, so the population is 800 × 2³. Since 2³ = 8, the population after 9 hours is 800 × 8 = 6,400. Adding 100% growth three times instead of compounding it — treating the growth as simple, not repeated doubling — gives 800 × 4 = 3,200, which is wrong because each period doubles the CURRENT population, not the original one. Using 9 as the power instead of dividing by the 3-hour period first gives 800 × 2⁹ = 409,600, which is wrong because the exponent counts periods, not hours. Giving the growth factor 2³ = 8 on its own, without multiplying by the starting population 800, leaves the answer as 8, which is wrong because the question asks for the population, not the multiplier. Always check that your final number of periods matches the total time divided by the period length.
- (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.
- (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) 118 m — Width = 840 ÷ 35 = 24 m. Perimeter = 2 × (length + width) = 2 × (35 + 24) = 2 × 59 = 118 m. The option 59 m gives the sum of the length and width but forgets to double it for the perimeter. The option 70 m doubles only the length (2 × 35 = 70) and leaves out the width entirely. The option 48 m doubles only the width (2 × 24 = 48) and leaves out the length entirely.
- (d) 8 — Working backwards by halving (undoing the doubling), one 4-hour step at a time: 9,600 (20 h) → 4,800 (16 h) → 2,400 (12 h) → 1,200 (8 h) → 600 (4 h) → 300 (0 h). Reading these in time order — 300, 600, 1,200, 2,400, 4,800, 9,600 at 0, 4, 8, 12, 16, 20 hours — the population is still at or below 1,000 at 4 hours (600) and first goes above 1,000 at 8 hours (1,200). So the recorded population first exceeds 1,000 at 8 hours. Answering 20 just reads off the time stated in the question, without working out when the threshold was actually first crossed — wrong, because 9,600 is only the value AT 20 hours, not necessarily the first time the population passed 1,000. Answering 0 comes from recovering the starting population by dividing 9,600 by 5 (the number of 4-hour gaps up to 20 hours) instead of by 2⁵ = 32 (the correct number of halvings), giving a wrongly-inflated starting value of 9,600 ÷ 5 = 1,920 — already above 1,000 at 0 hours — wrong, because doubling means the value must be halved five times, dividing by 2 five times (2⁵ = 32), not divided once by the number of gaps. Answering 12 comes from halving back only twice, from 9,600 to 4,800 (16 h) to 2,400 (12 h), and stopping there because 2,400 is already above 1,000, without checking that 1,200 at 8 hours is also above 1,000 and occurs earlier — wrong, because the FIRST recorded time above 1,000 is the earliest such time, not the first one reached while working backwards from 20 hours.
- (b) 1.50 — Substituting (0, 8) gives A = 8, since r⁰ = 1. Substituting (2, 18) gives 8 × r² = 18, so r² = 18 ÷ 8 = 2.25. Taking the square root of 2.25 gives r = 1.50 (2 d.p.). Stopping after finding r² and giving 2.25 as the final answer, without taking the square root, is wrong because r² is not the same as r. Dividing r² by the exponent 2 instead of taking its square root — treating the power as something you divide by rather than root — gives 2.25 ÷ 2 = 1.13 (2 d.p.), which is wrong. Inverting the ratio, working out 8 ÷ 18 instead of 18 ÷ 8, gives 0.44 (2 d.p.), which is wrong because the LATER value must be divided by the EARLIER one to find the growth multiplier, not the other way round.
- (c) 307.1 g — The multiplier for one hour is 1 − 0.15 = 0.85. After 3 hours the mass is 500 × 0.85³. Since 0.85 × 0.85 = 0.7225 and 0.85 × 0.7225 = 0.614125, the mass is 500 × 0.614125 = 307.0625 g, which rounds to 307.1 g. Stopping after only 2 hours instead of 3 gives 500 × 0.7225 = 361.25 g, which rounds to 361.3 g — wrong, because the question asks for 3 hours, not 2. Treating the decay as simple (15% × 3 = 45% lost in total, applied once) gives 500 × 0.55 = 275.0 g, which is wrong because the decay compounds hour by hour rather than adding up. Working out the AMOUNT LOST instead of the mass remaining gives 500 − 307.0625 = 192.9375 g, which rounds to 192.9 g — wrong, because the question asks what remains, not what has decayed away. Whenever a percentage decreases repeatedly, multiply by the same factor each period rather than adding the percentages together.
- (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) 4 years — The yearly multiplier is 1 − 0.12 = 0.88. After 1 year: 2400 × 0.88 = 2112. After 2 years: 2112 × 0.88 = 1858.56. After 3 years: 1858.56 × 0.88 = 1635.5328. After 4 years: 1635.5328 × 0.88 = 1439.27 (2 d.p.), which is the first value below £1,500 — so it takes 4 complete years. Stopping after 3 years, where the value is still £1,635.53, above £1,500, gives 3 years — wrong, because the threshold has not been crossed yet at that point. Continuing for one year longer than needed gives 5 years — wrong, because the value has already fallen below £1,500 by year 4, so year 5 overcounts. Multiplying by the percentage itself instead of by the multiplier — using 0.12 in place of 0.88 — gives 2400 × 0.12 = £288 after a single year, apparently below £1,500 straight away, so 1 year — wrong, because 12% is the part lost each year, and the part left is 88%, so the multiplier is 0.88.
- (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.
- (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.
- (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) 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.
- (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.
- (c) 16 km/h — Method: total distance = 12 + 8 = 20 km. Total time = 30 + 30 + 15 = 75 minutes = 1.25 hours. Average speed = total distance ÷ total time = 20 ÷ 1.25 = 16 km/h. Distractor origins: 20 km/h gives the total distance without ever dividing by the total time; 24 km/h uses only the speed of the first stage (12 km in 30 minutes), ignoring the second stage and the rest; 28 km/h averages the two separate stage speeds, 24 km/h and 32 km/h, instead of using total distance over total time.
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