Practise Gradients and areas under graphs. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
Ben cycled from home to a museum, stopped there for a while and then cycled back home. The distance–time graph shows the journey.
[object Object]
(a) How long did Ben stop at the museum? Give your answer in minutes.[1]
(b) Work out Ben's average speed on the way to the museum. Give your answer in km/h.[2]
(c) Was Ben cycling faster on the way to the museum or on the way back home? Give a reason for your answer.[1]
Show the answer and mark scheme
(a)Answer: 90 minutes
B1 for 90
Worked solution: The horizontal section lasts 1.5 hours = 90 minutes.
(b)Answer: 12 km/h
M1 for 18 ÷ 1.5
A1 for 12
Worked solution: \(\text{speed} = \frac{\text{distance}}{\text{time}} = 18 \div 1.5 = 12\) km/h
(c)Answer: on the way back home: the line is steeper
C1 for on the way back home with a reason, e.g. the line is steeper, or speeds 12 km/h and 18 km/h
Worked solution: Speed out \(= 12\) km/h, speed back \(= 18 \div 1 = 18\) km/h, so Ben was faster on the way back home (steeper line).
Question 2Medium5 marks
The graph gives information about the speed of a car during the first 60 seconds of a journey.
[object Object]
(a) Work out the acceleration of the car during the first 10 seconds.[2]
(b) Mo says, 'The car travelled more than 1 km during these 60 seconds.' Is Mo correct? You must show your working.[3]
Show the answer and mark scheme
(a)Answer: 2 m/s²
M1 for \(20 \div 10\)
A1 for 2
Worked solution: Acceleration = gradient \(= \frac{20}{10} = 2\) m/s2.
(b)Answer: No: distance = area under graph \(= 100 + 600 + 200 = 900\) m, which is less than 1 km (1000 m).
M1 for the area of one section, e.g. \(\frac{1}{2} \times 10 \times 20\) (= 100), \(30 \times 20\) (= 600) or \(\frac{1}{2} \times 20 \times 20\) (= 200)
M1 for a complete method for the total area, e.g. \(\frac{1}{2}(30 + 60) \times 20\) (= 900)
C1 for 'no' with 900 (m) compared with 1000 m
Worked solution: Distance = area under the graph \(= \frac{1}{2} \times 10 \times 20 + 30 \times 20 + \frac{1}{2} \times 20 \times 20 = 100 + 600 + 200 = 900\) m. 900 m < 1 km, so Mo is not correct.
Question 3Hard6 marks
A train travels between two stations. The graph shows its speed, \(v\) m/s, \(t\) seconds after it leaves the first station.
[object Object]
(a) Use 4 strips of equal width to work out an estimate for the distance between the two stations.[3]
(b) Is your answer to part (a) an underestimate or an overestimate of the actual distance? Give a reason for your answer.[1]
(c) Work out an estimate for the acceleration of the train at \(t = 50\).[2]
Show the answer and mark scheme
(a)Answer: 3750 m
M1 for reading the speeds at \(t = 50\), 100 and 150 (22.5, 30, 22.5; allow ±0.5)
M1 for the area of 4 trapezia (or triangles and trapezia) with width 50, e.g. \(\frac{1}{2} \times 50 \times (0 + 22.5) + \frac{1}{2} \times 50 \times (22.5 + 30) + \ldots\)
A1 for 3700 to 3800
Worked solution: Speeds: 0, 22.5, 30, 22.5, 0 at \(t = 0, 50, 100, 150, 200\). Area \(\approx \frac{1}{2} \times 50 \times [0 + 2(22.5 + 30 + 22.5) + 0] = 25 \times 150 = 3750\) m.
(b)Answer: Underestimate: the curve lies above the straight tops of the strips, so the strips miss some of the area under the curve.
C1 for 'underestimate' with a correct reason (the curve is above the tops of the trapezia / there are gaps between the strips and the curve)
Worked solution: The graph curves outwards (it is above each chord), so each trapezium is a little smaller than the area under the curve: the estimate is too small.
(c)Answer: 0.3 m/s² (accept 0.25 to 0.35) m/s²
M1 for a tangent drawn at \(t = 50\) and a correct method for its gradient
A1 for an answer from 0.25 to 0.35
Worked solution: Draw the tangent at \(t = 50\). It passes close to \((0, 7.5)\) and \((100, 37.5)\), so the gradient is about \(\frac{30}{100} = 0.3\) m/s2.