The diagram shows the velocity-time graph for the motion of a bus [Figure 3.1]. The bus starts from rest and accelerates uniformly for 8 seconds until it reaches a speed of 12.6ms¯¹. The bus maintains this speed for 40 seconds. It then decelerates uniformly in two stages. Between 48 and 62 seconds the bus decelerates at am s¯² and between 62 and 70 seconds it decelerates at 2ams¯² until coming to rest.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

1. [B1] (a) Distance = 50.4m. Allow 252/5.
2. [M1] (b) ν₁ = 12.6-(62-48)a. Use of suvat for first section of deceleration. 12.6±(62-48)a only.
3. [M1] (b) 0=v₁-2ax(70-62). Use of suvat for second section of deceleration. An expression for the velocity at 62 seconds must be ±2ax(70-62).
4. [A1] (b) a=0.42. -0.42 scores A0.
5. [B1] (c) Speed at time t=62 is 6.72ms⁻¹. This may be seen in part (b) but must be used in part (c) to get this mark.
6. [B2FT] (c) S2 = (48-8)×12.6 [= 504]. S3 = 0.5×(12.6+ their 6.72)×(62-48) [ = 135.24or 3381/25 ое ] or their 6.72×(62-48) +0.5×(62-48)×(12.6-their 6.72). S4 = 0.5× their 6.72×(70-62) [= 26.88 or 672/25 ое]. B2 FT for any 2 correct, B1 FT for any 1 correct – follow through their value of v₁ where 0<v₁ <12.6 but must have come from the correct equations seen in part (b). Allow correct value of v₁ from a=-0.42 where v₁ = 12.6+(62-48)a and v₁ = −2a×(70-62).
7. [B1] (c) Average speed =10.236 ms⁻¹. Allow 10.2 or better oe e.g. 2559/250, 59/10, 250.