The speed of a bullet is measured to be ${}^{\mathbf{640}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$ as the bullet emerges from a barrel of length ${}^{\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}}$. Assuming constant acceleration, find the time that the bullet spends in the barrel after it is fired.

The final speed of the bullet,${}^{\mathbf{v}\mathbf{=}\mathbf{}\mathbf{640}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$

The length of the barrel, ${}^{\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}}$

Kinematic equations describe the motion of an object with constant acceleration. These equations can be used to determine the acceleration, velocity or distance.

The expression for the kinematic equations of motion are given as follows:

$\mathbf{v}\mathbf{=}\mathbf{}{\mathbf{v}}_{\mathbf{0}\mathbf{}\mathbf{+}\mathbf{}\mathbf{at}}$ … (i)

${\mathbf{v}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{ax}$ … (ii)

Here, ${}^{{\mathbf{v}}_{\mathbf{0}}}$ is the initial velocity, ${}^{\mathbf{v}}$is the final velocity, ${}^{\mathbf{t}}$is the time, ${}^{\mathbf{a}}$is the acceleration and${}^{\mathbf{x}}$is the displacement.

Since the bullet starts from rest, the initial speed is zero, that is,${}^{{\mathbf{v}}_{\mathbf{0}}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$ .

Using equation (ii), the acceleration can be calculated as follows:

$$\begin{gathered} \mathbf{a}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}\mathbf{-}{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}}}{\mathbf{2}\mathbf{x}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{{\left(640m/s\right)}^{\mathbf{2}}\mathbf{}\mathbf{-}{\left(0m/s\right)}^{\mathbf{2}}}{\mathbf{2}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}} \end{gathered}$$

Using equation (i), the time spent by the bullet is calculate as follows:

$$\begin{gathered} \mathbf{t}\mathbf{=}\frac{\mathbf{v}\mathbf{-}{\mathbf{v}}_{\mathbf{0}}}{\mathbf{a}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{640}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}\mathbf{-}\mathbf{}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{s}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{76}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s} \end{gathered}$$

Therefore, the time spent by the bullet in the barrel after it is fired is${}^{\mathbf{3}\mathbf{.}\mathbf{76}\mathbf{}\mathbf{ms}}$ .