College Physics by Openstax Chapter 2 Problem 22

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=6.20×105 m/s2a=6.20 \times 10^{5} \ \text{m/s}^2; Δt=8.10×10−4 s\Delta t=8.10 \times 10^{-4} \ \text{s}; and v0=0m/sv_0=0 \text{m/s}.

The muzzle velocity of the bullet is computed as follows:

vf=v0+atvf=0 m/s+(6.20×105 m/s2)(8.10×10−4 s)vf=502 m/s  (Answer)\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. 

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