Question

1. a bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20×10^5 m/s² for 8.10×10^(-4) s. what is its muzzle velocity (that is, its final velocity)?

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v = v_0+at$. Since the bullet starts from rest, $v_0 = 0$.

Step2: Substitute the values

Given $a = 6.20\times10^{5}\ m/s^{2}$ and $t = 8.10\times10^{-4}\ s$. Substituting into $v=v_0 + at$ (with $v_0 = 0$), we get $v=at$.
$v=(6.20\times 10^{5}\ m/s^{2})\times(8.10\times 10^{-4}\ s)$

Step3: Calculate the result

$v = 6.20\times8.10\times10^{5}\times10^{-4}\ m/s$
$v = 50.22\times10^{1}\ m/s$
$v = 502.2\ m/s$

Answer:

$502.2\ m/s$