A particle moves in a straight line starting from a point O before coming to instantaneous rest at a point X. At time t s after leaving O, the velocity v ms¯¹ of the particle is given by v = 7.2t² 0 ≤ t ≤ 2, v = 30.6 – 0.9t 2 ≤ t ≤ 8, v = 1600/t² + kt 8 ≤ t, where k is a constant. It is given that there is no instantaneous change in velocity at t = 8. Find the distance OX.

Mark scheme

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

1. [*M1] Use velocity at t = 8 to set up a linear equation in k only. Allow a slip in one value or sign only. 30.6-0.9×8=1600/8² +8k
2. [A1] k =-0.2
3. [DM1] Attempt to find the value of t when the particle comes to rest using the correct expression for v, set equal to zero with their negative value of k. Must find a positive value for t (for reference, t = 20).
4. [*M1] Attempt to integrate v for one of the 3 intervals. Increase power by 1 and a change in coefficient in at least one term (which must be the same term); s = vt is M0.
5. [A1] S=7.2t³/3 (+c). May be unsimplified (for reference, limits are from 0 to 2).
6. [A1] S = 30.6t - 0.9t²/2 (+c). May be unsimplified (for reference, limits are from 2 to 8).
7. [A1FT] S = -1600/t - 0.1t² (+c). May be unsimplified (for reference limits are from 8 to 20). Follow through their value of k or just k only.
8. [B1] One correct distance found. Allow unsimplified e.g. (216-59.4) or (1/2)×(8-2)×(28.8+23.4) etc.
9. [B1] Distance =19.2+(216-59.4)+(-120-(-206.4)) = 262.2 m. This mark can be awarded if no integration is shown oe. e.g. 1311/5. Condone 262 www.