vf^2=vi^2+2ad
solve for a
solve for a
\(v^2 = u^2 + 2as\)
where:
- \(v\) is the final velocity of the bullet (which is the muzzle velocity in this case),
- \(u\) is the initial velocity of the bullet (which is zero because the bullet starts from rest in the barrel),
- \(a\) is the acceleration of the bullet (which is what we need to find),
- \(s\) is the distance traveled by the bullet in the barrel.
Plugging in the given values, we get:
\(v^2 = u^2 + 2as \)
\( (553 m/s )^2 = (0 m/s)^2 + 2a(0.648 m) \)
\( 305609 m^2/s^2 = 2a(0.648 m) \)
Now, we can solve for the acceleration (\(a\)) by isolating it in the equation:
\( a = \frac{{305609 m^2/s^2}}{{2 \times 0.648 m}} \)
Calculating this, we find:
\( a \approx 237053.4 m/s^2 \)
Therefore, the acceleration of the bullet is approximately \(237053.4 m/s^2\).