M1 acceleration graph question

A car is travelling along a straight horizontal road. The car takes 120 s to travel between two sets of traffic lights which are 2145 m apart. The car starts from rest at the first set of traffic lights and moves with constant acceleration for 30 s until its speed is 22 m s–1. The car maintains this speed for T seconds. The car then moves with constant deceleration, coming to rest at the second set of traffic lights. (a) Sketch, in the space below, a speed-time graph for the motion of the car between the two sets of traffic lights. (2)

(b) Find the value of T. (3) A motorcycle leaves the first set of traffic lights 10 s after the car has left the first set of traffic lights. The motorcycle moves from rest with constant acceleration, a m s–2, and passes the car at the point A which is 990 m from the first set of traffic lights. When the motorcycle passes the car, the car is moving with speed 22 m s–1.
(c) Find the time it takes for the motorcycle to move from the first set of traffic lights to the point A. (4)
(d) Find the value of a.

Hi, could someone help with with part d please? I have done abc and found that it takes 50s for the motorcyle to catch up with the car. However for part d , why can't do you the equation a=(v-u)/2 ? The equation that the mark scheme has included t and d. But surely a=(v-u)/2 would give me the same thing since both are used to find a. Is this to do with a being constant? Thanks.

I don't think that equation exists- do you mean a=v-u/t instead of 2?

That equation doesn't have a in it though :/
isn't the correct version of that $s=\left(\dfrac{u+v}{2}\right)t$?

Oh yeah, sorry that's the one I was talking about a=(v-u)/t. Thanks.

No problem, I always used to get 2 and T mixed up too, they sound so similar in my head lol

That equation doesn't have a in it though :/
isn't the correct version of that $s=\left(\dfrac{u+v}{2}\right)t$?

this equation does exist and its right but we hardly use it in M1

hardly yes.... but its still there... and i'd rather use it than $s=ut+\frac{1}{2} at^2$

and $v^2 = u^2 +2as$