A bullet leaves a rifle barrel of length l = 0.6 m with a…

A bullet leaves a rifle barrel of length $l = 0.6\ \mathrm{m}$ with a speed $v = 600\ \mathrm{m/s}$. How long does the bullet spend in the barrel?

Explanation of Method

Time a bullet spends in the barrel.

Step 1: Assumptions

The bullet starts from rest ($v_0 = 0$) and speeds up uniformly (constant acceleration) along the barrel.

Step 2: Use the average speed

For uniform acceleration, the average speed is the average of the initial and final speeds: \[v_{\text{avg}} = \frac{v_0 + v}{2} = \frac{0 + 600}{2} = 300\ \mathrm{m/s}.\]

\[ t = \frac{l}{v_{\text{avg}}} = \frac{2l}{v} = \frac{2 \cdot 0.6\ \mathrm{m}}{600\ \mathrm{m/s}}. \]

\[t = 0.002\ \mathrm{s} = 2\ \mathrm{ms}\]

$v\ (\mathrm{m/s})$ $t\ (\mathrm{ms})$ $0$ $1$ $2$ $300$ $600$ $\text{area} = l = 0.6\ \mathrm{m}$ $v_{\text{avg}} = 300\ \mathrm{m/s}$

Inside the barrel the speed grows in a straight line from 0 to 600 m/s. The area under the line is the barrel length; the dashed average speed 300 m/s over the same time gives the same area.

Step 3: The acceleration (optional)

\[ a = \frac{v^2}{2l} = \frac{600^2}{2 \cdot 0.6} = 3 \cdot 10^5\ \mathrm{m/s^2}, \] about 30 000 times the acceleration due to gravity.

Check

$ v = a t = 3 \cdot 10^5\ \mathrm{m/s^2} \cdot 0.002\ \mathrm{s} = 600\ \mathrm{m/s}\ \checkmark $

$ l = \tfrac12 a t^2 = \tfrac12 \cdot 3 \cdot 10^5 \cdot (0.002)^2 = 0.6\ \mathrm{m}\ \checkmark $

Result

\[\boxed{\,t = 0.002\ \mathrm{s} = 2\ \mathrm{ms}\,}\]

Remark

Note. The original solution first found $a$ and then $t = v/a$; the shortcut $t = 2l/v$ gives the same result in one step and does not need $a$ at all.