91Ó°ÊÓ

Initial speed of muon is,$v_i=5.00×{10}^6\mathrm{m}/\mathrm{s}$

Acceleration of the muon is,$a=-1.25×{10}^{14}\mathrm{m}/{\mathrm{s}}^2$

Kinematic equations of motion describe the relationship between displacement, velocity or acceleration of a particle. The graphs of displacement vs time and velocity vs timecan be plotted to analyze the motion of the particle.

The expression for the equation of motion is given as below,

$v_f^2=v_i^2+2ax$(i)

Here, $v_f$is the final velocity,$v_i$ is the initial velocity, $a$is the acceleration, and $x$is the distance.

Rearrange the equation (i) for x and substitute the values.

$$\begin{gathered} x=\frac{v_f^2-v_i^2}{2a} \\ =\frac{\left(0^2-{\left(5.00×{10}^6\right)}^2\right)}{\left(2×\left(-1.25\right)×{10}^{14}\right)} \\ =0.100\mathrm{m} \end{gathered}$$

To plot graph of vs kinematic equationis used,

$$\begin{gathered} x=v_{i}t+\frac{1}{2}a{t}^2 \\ x=\left(5.00×{10}^6×t\right)+\left(\frac{1}{2}×\left(-1.25×{10}^{14}×t^2\right)\right) \end{gathered}$$

To plot graph of vs t kinematic equation is used,

$$\begin{gathered} v=v_i+at \\ v=\left(5.00×{10}^6\right)+\left(-1.25×{10}^{14}×t\right) \end{gathered}$$