When braking, the car moves with equal deceleration along the highway with a rounding of radius R

When braking, the car moves with equal deceleration along the highway with a rounding of radius R and goes to a complete stop along the path S, having an initial speed V0. Determine the time of movement of the car to a complete stop and calculate the values of full accelerations at the initial moment and after time t1. R, = 100 S, = 70 V0 = 36 t1 = 4

We use the formula: S = (V ^ 2-V0 ^ 2) / 2a where a is the acceleration, then
S * 2a = V ^ 2-V0 ^ 2
a = (V ^ 2-V0 ^ 2) / 2S = -36 ^ 2/2 * 100 = -6.48 m / s ^ 2
Since S = at ^ 2/2, the time to a complete stop is t = √2S / a = √140 / 6.48 = 4.65 s.

The speed changes according to the law v (t) = v0 + at then v (4) = 36-6.48 * 4 = 10 m / s
Since the car moves in a circle, a centripetal force acts on it, ac = v ^ 2 / r = 10 ^ 2/100 = 1 / s ^ 2
The total acceleration A will be equal to √ (a ^ 2 + ac ^ 2) = √6.48 ^ 2 + 1 = 6.56 m / s ^ 2