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.