Question

A cannon ball barrel is 6m long. The muzzle velocity of a bullet from the cannon is 305 m/s. What is the average acceleration of the bullet going through the bullet?

Answer 1

Assuming that the acceleration is constant will write the equation for displacement (1) and acceleration (2):

$s = s_0 + v_0t + \cfrac{at^2}2 \qquad \text{(1)} \\ a = \cfrac{v-v_0}t \qquad \text{(2)}$

Сonsidering that the initial velocity ($v_0$ ) and initial displacement ($s_0$ ) are zero, can write:

$s = \cfrac{at^2}2 \qquad \text{(1.a)} \\ a = \cfrac v t \qquad \text{(2.a)}$

Multiply (1.a) of $2a$ :

$2as = a^2t^2 \qquad \text{(1.b)} \\ a = \cfrac {v} {t} \qquad \text{(2.b)}$

Substitute the expression for $a$ from (2.b) into the right side of (1.b)

$2as = \Big(\cfrac v t\Big)^2 t^2 = v^2 \qquad \text{(1.c)}$

Define the acceleration:

$a = \cfrac {v^2} {2s} = \cfrac {305^2} {2*6} \approx 7752 \text{ (m/s)}$