Question

A muon (an elementary particle) enters a region with a speed of$\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$and then is slowed down at the rate of$\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. (a)How far does the muon take to stop? (b) Graph x vs t and v vs t for the muon.

Short answer

(a) The distance muon takes to stop is 0.100 m .

(b) The graphs of x vs t and v vs tare plotted for the muon.

Step-by-step solution

Given Data

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

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

Understanding the kinematic equations of motion.

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.

(a) Determination of the distance required for muon to stop

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

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

(b) Plotting the graph of x vs t and v vs t.

To plot graph of vs kinematic equationis used,

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

To plot graph of vs t kinematic equation is used,

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