O-LevelMathematicsAlgebra and graphs4024/22 · May/June 2022 · Q710 Marks

[Figure with a speed-time graph] The diagram shows the speed-time graph for a cyclist's journey.

(a)(i)Calculate the acceleration of the cyclist during the first 8 seconds.[1]
(a)(ii)Describe the motion of the cyclist between t = 8 and t = 90.[1]
(a)(iii)The total distance travelled by the cyclist during the journey is 558 m. Find the value of T.[3]
(a)(iv)Convert 6 m/s into km/h.[2]
(b)A car travels 352km, correct to the nearest kilometre. The time taken to travel this distance is 4.2 hours, correct to the nearest 0.1 hour. Calculate the upper bound for the average speed of the car.[3]

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Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

[1] 0.75 or 3/4 or 6/8
[1] Travelling at a constant/uniform speed oe or Travelling at acceleration of 0 oe
[3] M2 for 1/2 × 6 × (T + (90–8)) [=558] oe or (T-90) × 6 = 42 oe or 1/2 × 6 × ((90-8)+T – 8) = 534 oe or M1 for 1/2 × 8 × 6 oe or (90 – 8) × 6 oe or 1/2 (90+(90-8))×6 oe seen After 0 scored, SC1 for answer 14
[2] B1 for answer figs 216 or M1 for 6/1000 x 60 [x60] oe
[3] B1 for 352.5 or 4.15 seen M1 for their352.5 / their4.15

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