# What is the acceleration of a bullet that, having broken through a 35cm thick wall, reduced its velocity

**What is the acceleration of a bullet that, having broken through a 35cm thick wall, reduced its velocity from 800m / s to 400m / s?**

Given:

S = 35 centimeters = 0.35 meters – the thickness of the wall pierced by the bullet;

V0 = 800 m / s – initial bullet velocity;

V1 = 400 m / s – final bullet velocity.

It is required to determine a (m / s ^ 2) – bullet acceleration.

The bullet has decreased its speed over time:

t = (V0 – V1) / a;

Then:

S = V0 * t – a * t ^ 2/2 = V0 * (V0 – V1) / a – a * (V0 – V1) ^ 2 / (2 * a ^ 2) =

V0 * (V0 – V1) / a – (V0 – V1) ^ 2 / (2 * a) = (2 * V0 * ((V0 – V1) – (V0 – V1) ^ 2) / (2 * a) =

= (V0 – V1) * (2 * V0 – V0 + V1) / (2 * a) = (V0 – V1) * (V0 + V1) / (2 * a), hence:

a = (V0 – V1) * (V0 + V1) / (2 * S) = (800 – 400) * (800 + 400) / (2 * 0.35) =

= 400 * 1200 / 0.7 = 685614 m / s ^ 2.

Answer: bullet acceleration is 685 km / s ^ 2.