How do I answer this mechanics question on distance?

Your answer for the final velocity after 12 seconds seems to be correct.

use the equation v^2 = u^2 + 2as and rearrange for s.

Thank you I got the correct answer using this formula. But I wrote the acceleration as -1 since it's decelerating. However, why is my distance negative? As in -162m instead of 162.

I am also confused as to why I'm getting wrong answers using the other formulae? I tried s = ut + 1/2at^2, I did (0x12) +1/2(-1)(144) ended up with -72, and s = 1/2 (u+v)t , I did 1/2 (0 + 18)x 12 ended up with 108m. Also s = vt - 1/2at^2 , (18 x 12) - 1/2 (-1x 144) and got 288m.

u is not zero in the last part of the motion.

u = 18 (initial velocity when start breaking)
v =0 (when car is at rest).

You don't need to use the equation: s=ut +1/2 at^2 since you are not given time. 12 seconds is the time it takes for the car to accelerate up to 18 ms^-1, not the time it takes to decelerate to 0 ms^-1. If you want to use the equation s = ut + 1/2at^2, you need to find the time it takes to decelerate to 0 ms^-1 from 18 ms^-1 by using use v = u + at.

Thank you for the reply, so u = 18 and v = 0? In that case, when I do (18x12) + 1/2 (-1x144), I got 144m. And (0 x 12) - 1/2 (-1 x 144) gets me 72m. and 1/2 (18 + 0) x 12 = 108m??

Where does the 12 come from?
in the second part of the motion u = 18 v = 0 a = -1

I see, time isn't given so I needed to figure it out first

Is time needed to get s?

No, i just wanted to practice understanding how to use all the formulae i could, i got 18s for the new time and 162m using them

A car starts from rest and accelerates constantly at 1.5 ms^-2. After 12 seconds the driver breaks causing the car to decelerate at a constant rate of 1 ms^-2.

Calculate the distance the car travels from the instant the driver brakes until the car comes to rest.

I found the final velocity to be 18 m/s after 12 seconds. But I keep getting 108 m for distance I used s = (0 + 18)/2 x 12. What do I do??