A muon (an elementary particle) enters a region with a speed of5.00×106m/sand then is slowed down at the rate of1.25×1014m/s2. (a)How far does the muon take to stop? (b) Graph x vs t and v vs t for the muon.

Short Answer

Expert verified

(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

01

Given Data

Initial speed of muon is,vi=5.00×106m/s

Acceleration of the muon is,a=-1.25×1014m/s2

02

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,

vf2=vi2+2ax(i)

Here, vfis the final velocity,vi is the initial velocity, ais the acceleration, and xis the distance.

03

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

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

x=vf2-vi22a=02-5.00×10622×-1.25×1014=0.100m

04

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

To plot graph of vs kinematic equationis used,

x=vit+12at2x=5.00×106×t+12×-1.25×1014×t2

To plot graph of vs t kinematic equation is used,

v=vi+atv=5.00×106+-1.25×1014×t

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