Velocity at time T is 2T soi
21(8+4−T)×2T=27
M1 for an attempt at trapezium area or three areas and =27 using their 2T.
(T-3)(T-9)=0
M1 dep on previous M1 for solving their 3-term quadratic by factorising, completing the square, or quadratic formula.
T=3
Alternative method:
Velocity at time T is 2T, leading to T=2v soi
21(8+4−2v)×v=27
(M1) For an attempt at trapezium area =27 using their 2v.
(18-v)(v-6)=0
(M1) Dep on previous M1 for solving their 3-term quadratic.
v=6, then T=3