A car is travelling along a straight horizontal road. The car passes through a point A, on the road travelling at a speed of 15ms⁻¹, and then accelerates uniformly at 0.4ms⁻² for 30 seconds. The car then moves at constant speed for 37 seconds, where T < 30. The car then decelerates uniformly at 0.2 ms⁻² and after a further T seconds passes through a point B on the road. [Figure 4.1]

Mark scheme

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

1. [NOTE] The mark scheme for this question was not used by examiners. All candidates were awarded full marks for this question due to a series-specific issue.
2. [1] (a) Correct three-line segments for v-t graph First line segment with positive gradient starting on the positive vertical axis, second line segment horizontal (parallel to t -axis) and third line segment with negative gradient, stopping before the horizontal axis. If the third line continues and touches the t -axis, there must be some indication that speed of the car at B is not 0.
3. [1] (a) Values/expressions labelled on axes correctly 15 correct on vertical axis; 30, 30+ 3T and 30 + 4T correctly labelled on horizontal axis.
4. [1] (b) Speed after 30 seconds = 15+0.4×30[=27]
5. [1] (b) Speed at B is their 27+(-0.2)T Use of v=u+at using their speed after 30 seconds for u, t = T and a =±0.2.
6. [1] (b) Attempt at distance from A to B and equate to 2750 Using the combined area below the three line segments; using their 27.
7. [1] (b) 1/2(15+27)×30+27×3T+1/2{(27-0.2T)+27}×T = 2750 Correct (un-simplified) equation for T.
8. [1] (b) 0.1T²-108T+2120=0⇒T=... Re-arranging and attempting to solve their three-term quadratic equation in T. If method seen, must be using correct formula OR if factorising, two terms must be correct for their three-term quadratic when expanding brackets. If no method seen, must have at least one correct value for their three-term quadratic for this mark.
9. [1] (b) T = 20 only If T=1060 also stated, then must be rejected.
10. [1] (c) Speed at B is (their 27)+(-0.2)×(their 20)[=23] Use of v = u + at using their speed after 30 seconds from (b), their T from (b) and a = ±0.2.
11. [1] (c) Distance from B to C is s where 0² = (their 23)²+2x(-0.5)×s and attempt to solve for s Use of v² =u²+2as with v=0 and a = -0.5 and their speed at B for u (if correct s = 529).
12. [1] (c) Total distance is [529 + 2750 =] 3279 m Condone 3280 m.