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A muon (an elementary particle) is shot with initial speed 5.20 X 106 m/s into a region where an electric field produces an acceleration of 1.30 X 1014 m/s? directed opposite to the initial velocity. How far does the muon travel before coming to rest?

Navjyot Kalra
6 years ago
We assume that the motion is along the x direction only.
Given:
Initial speed of muon, v0x = 5.20 x 106.
Final speed of muon, vx = 0 m
Acceleration of muon, ax = -1.30 x 1014 m/s2.
It is important to note that the negative sign of the acceleration is due to the fact that the direction of the acceleration is opposite to that with the direction of initial velocity.
Let us assume that the time taken by the muon to come at rest, under the action of acceleration is given by t. Therefore the time t can be given as:

With initialcondition x0 = 0, xis given as:

Substitute the given values to obtain the value of x as:

Therefore the muon has gone 0.104 m from the starting point.
11 Points
3 years ago
Work=  force*distance
distance=work/force
here work is nothing but the kinetic energy of moving muon in the initial direction. and force is mass times acceleration produced.
distance= (1/2*m*v​2​)/ma