A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20×10⁵ m/s² for 8.10×10^-4 s . What is its muzzle velocity (that is, its final velocity)?

Solution:

We are given the following: a=6.20 \times 10^{5} \ \text{m/s}^2; \Delta t=8.10 \times 10^{-4} \ \text{s}; and v_0=0 \text{m/s}.

The muzzle velocity of the bullet is computed as follows:

\begin{align*}
v_f & =v_0+at \\
v_f & = 0\:\text{m/s}+\left(6.20\times 10^5\text{ m/s}^2\right)\left(8.10\times 10^{-4}\:\text{s}\right) \\
v_f & =502\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the muzzle velocity, or final velocity, is 502 m/s.