• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q26P

Expert-verified
Found in: Page 34

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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

(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.

See the step by step solution

Step 1: Given Data

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

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

Step 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.

Step3: (a) Determination of the distance required for muon to stop

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

$x=\frac{{v}_{f}^{2}-{v}_{i}^{2}}{2a}\phantom{\rule{0ex}{0ex}}=\frac{\left({0}^{2}-{\left(5.00×{10}^{6}\right)}^{2}\right)}{\left(2×\left(-1.25\right)×{10}^{14}\right)}\phantom{\rule{0ex}{0ex}}=0.100\mathrm{m}$

Step 4: (b) Plotting the graph of x vs t and v vs t.

To plot graph of vs kinematic equationis used,

$x={v}_{i}t+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}x=\left(5.00×{10}^{6}×t\right)+\left(\frac{1}{2}×\left(-1.25×{10}^{14}×{t}^{2}\right)\right)$

To plot graph of vs t kinematic equation is used,

$v={v}_{i}+at\phantom{\rule{0ex}{0ex}}v=\left(5.00×{10}^{6}\right)+\left(-1.25×{10}^{14}×t\right)$