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.The table shows the speed, in m/s, of a car at times, in seconds, all 2 seconds apart: 0 at t = 0, 3 at t = 2, 8 at t = 4, 15 at t = 6, 24 at t = 8. A student uses the trapezium rule with these five readings to estimate the distance the car travels. By finding the second differences of the speed values, decide whether this trapezium estimate is an overestimate or an underestimate of the true distance.
- 2.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.
- 3.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?
- 4.Two students each use the trapezium rule to estimate the area under the same curve between x = 0 and x = 8. Student A uses 4 strips of width 2. Student B uses 8 strips of width 1. Which estimate is more likely to be closer to the true area, and why?
- 5.On a distance-time graph, a horizontal line segment shows a period when the graph's gradient is 0. What does this tell you about the journey during that time?
- 6.A tram travels between two stops. Its velocity-time graph consists of straight line segments joining the points (0, 0), (5, 20), (12, 20), (16, 4) and (20, 4), where time is in seconds and velocity is in m/s. Work out the average speed of the tram over the whole 20 seconds. Give your answer to 1 decimal place.
- 7.A bath is filled with water; the graph of volume (litres) against time (minutes) is a curve, since the flow rate changes. The tangent to the curve at t = 10 minutes passes through the points (8, 130) and (12, 190), and the volume in the bath at t = 10 minutes is 160 litres. Use the gradient of this tangent to estimate the volume in the bath 5 minutes after t = 10 minutes.
- 8.A car's journey is represented by a velocity-time graph. The total distance travelled, found from the area under the graph, is 350 metres, and the total time for the journey is 25 seconds. Work out the car's average speed for the whole journey, in m/s.
- 9.The curve y = x² − 3x. Use the chord between x = 1 and x = 4 to estimate the gradient of the curve at x = 2.5.y = x² − 3x
- 10.A car's speed, in m/s, during a journey is: 0 at t = 0 s, 8 at t = 4 s, and 8 (constant) from t = 4 s to t = 10 s, increasing at a constant rate between t = 0 and t = 4. Estimate the total distance the car travels, using the areas of a triangle and a rectangle.
- 11.A cup of tea cools after being made. Its temperature-time graph is a curve. Three tangents are drawn on the graph: at t = 2 minutes the tangent has gradient −8; at t = 10 minutes the tangent has gradient −3; at t = 20 minutes the tangent has gradient −0.5. Which of these statements about the tea's cooling is correct?
- 12.The trapezium rule is used to estimate the area under a curve between x = 2 and x = 6. The curve is concave up (it curves upwards, like the inside of a bowl) throughout this interval. Which statement about the estimate is correct?
- 13.A tap's flow rate, in litres per minute, is plotted against time, in minutes, on a graph. What are the units of the area under this graph?
- 14.The area under a speed-time graph between t = 0 and t = 6 seconds is estimated using three strips of equal width, using the speeds, in m/s, at t = 0, 2, 4 and 6: 0, 5, 9 and 12. Using the trapezium rule with these three trapezia, estimate the distance travelled.
- 15.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.
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