A-LevelMathematicsMechanics9709/42 · Oct/Nov 2025 · Q38 Marks

A car of mass 1400 kg travelling on a straight horizontal road accelerates uniformly from a speed of 15 m s⁻¹ to 20 m s⁻¹ over a distance of 175 m.

(a)Find the time taken to travel the 175 m.[2]
(b)(i)Use an energy method throughout to find the average power of the car's engine as the speed increases from 15 m s⁻¹ to 20 m s⁻¹.[4]
(b)(ii)Find the steady speed that the car could maintain on the horizontal road if the engine is working at the power found in (b)(i).[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

[M1] For complete method to get an equation in time only using s=175, u=15 and v = 20 . May see a=0.5 .
[A1] Time = 10 s; CWO.
[B1] Change in KE = ±(½×1400×20² - ½×1400×15²) [=±(280000-157500)=±122500] OR Work done against resistance = ±800×175 [= ±140000]
[*M1] Work done by engine = ½×1400×20² - ½×1400×15² +800×175; Attempt at work energy equation. Correct number of relevant terms; dimensionally correct; allow sign errors. May see : Work done by engine = DF×175. Using ½×1400×(20-15)² is M0. For reference if correct WD=262500, DF=1500.
[DM1] Av power×(their 10) = ½×1400×20² - ½×1400×15² + 800×175; FT their 10 from 3(a) or if recalculated t in 3(b)(i)
[A1FT] Average power = 26250 W; FT their value of t only so 262500/their 10 or 1500×175/their 10. Allow 26300 from correct working. Allow 26.25 kW or 26.3 kW with units.
[M1] their 26250 = 800v; For use of P=Fv.
[A1] Speed = 32.8125 or 525/16 m s⁻¹; OE. Allow 32.875 or 32.9 or better from use of 26300.

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