When braking, the car, moving at uniform acceleration, in the fifth second of movement travels a distance

When braking, the car, moving at uniform acceleration, in the fifth second of movement travels a distance of 5 cm and stops. What way did the car go in 3 seconds, movements

S5 = 5 cm = 0.05 m.
t5 = 1 s.
t = 5 s.
t (3) = 3 s.
t (2) = 2 s.
S3 -?
Since the final speed is 0, the car stopped, then the path S5 in 5 (fifth second) can be expressed as S5 = a * t5 ^ 2/2, where a is the acceleration of the car, t5 is the travel time in 5 seconds.
a = 2 * S5 / t5 ^ 2.
a = 2 * 0.05 m / (1 s) ^ 2 = 0.1 m / s ^ 2.
Let’s find the speed at the beginning of braking Vo = a * t.
Vo = 0.1 m / s ^ 2 * 5 s = 0.5 m / s.
S3 = S (3) – S (2), where S (3) is the distance covered in three seconds of braking, S (2) is the distance covered in two seconds of braking.
S (3) = Vo * t (3) – a * t (3) ^ 2/2, where t (3) – movement time – three seconds.
S (3) = 0.5 m / s * 3 s – 0.1 m / s ^ 2 * (3 s) ^ 2/2 = 1.05 m.
S (2) = Vo * t (2) – a * t (2) ^ 2/2, where t (2) – movement time – two seconds.
S (2) = 0.5 m / s * 2 s – 0.1 m / s ^ 2 * (2 s) ^ 2/2 = 0.8 m.
S3 = 1.05 m – 0.8 m = 0.25 m.
Answer: in the third second of braking, the car covered the distance S3 = 0.25 m.