Question

a bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.40×10^5 m/s² for 7.30×10^(-4) s. what is its muzzle velocity (in m/s) (that is, its final velocity)? (enter the magnitude.)

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 $a = 6.40\times10^{5}\ m/s^{2}$ and $t=7.30\times 10^{-4}\ s$. Substitute into $v = v_0 + at$. Since $v_0 = 0$, we have $v=at$.
$v=(6.40\times 10^{5}\ m/s^{2})\times(7.30\times 10^{-4}\ s)$

Step3: Calculate the result

$v = 6.40\times7.30\times10^{5}\times10^{-4}\ m/s$
$v = 46.72\times10^{1}\ m/s$
$v = 467.2\ m/s$

Answer:

$467.2$