Question

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.

Short answer

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

Step-by-step solution

Given data

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

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

Understanding the kinematic equations

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.

Determination of the acceleration of bullet.

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:

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

Determination of the time spent by the bullet in barrel

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${}^{\mathbf{3}\mathbf{.}\mathbf{76}\mathbf{}\mathbf{ms}}$ .