# When braking, the car moves with equal deceleration and stops 30s after the start of braking, having covered

**When braking, the car moves with equal deceleration and stops 30s after the start of braking, having covered a distance of 120m. With what acceleration did the car move and what was its initial speed?**

V = 0 m / s.

t = 30 s.

S = 120 m.

V0 -?

a -?

The movement of the car S during uniformly accelerated braking is expressed by the formula: S = V0 * t – a * t ^ 2/2, where V0 is the initial speed of the car, t is the acceleration time, and is the acceleration during acceleration.

Let us express the acceleration of the car a during braking by the formula: a = (V0 – V) / t, where V0, V is the speed of the car at the beginning and end of braking.

Since the car stopped at the end of braking V = 0 m / s, the formula will take the form: a = V0 / t.

V0 = a * t.

S = a * t * t – a * t ^ 2/2 = a * t ^ 2/2.

a = 2 * S / t ^ 2.

a = 2 * 120 m / (30 s) ^ 2 = 0.27 m / s2.

V0 = 0.27 m / s2 * 30 s = 8.1 m / s.

Answer: before braking, the car was moving at a speed of V0 = 8.1 m / s, while braking, it was moving with an acceleration of a = 0.27 m / s2.