# The car was moving at a speed of 20 m / s, then braking occurred with an acceleration of 4 m / s2

**The car was moving at a speed of 20 m / s, then braking occurred with an acceleration of 4 m / s2. Find the distance covered by the car 6 seconds after the start of braking.**

V0 = 20 m / s.

a = 4 m / s2.

t = 6 s.

S -?

The displacement of the body S with equally slow motion is expressed by the formula: S = (V0 ^ 2 – V ^ 2) / 2 * a, where V0, V are the initial and final velocity of the body, and is the acceleration during deceleration.

We express the final velocity V of the body from the formula for the acceleration of the body: а = (V0 – V) / t.

V0 – V = a * t.

V = V0 – a * t.

V = 20 m / s – 4 m / s2 * 6 s = – 4 m / s.

We see that if the braking force acted t = 6 s, then the car should stop and start moving in the opposite direction.

Therefore, the final speed of the vehicle will be V = 0m / s.

S = V0 ^ 2/2 * a.

S = (20 m / s) ^ 2/2 * 4 m / s2 = 50 m.

Answer: the car went to a full stop the path S = 50 m.