A muon (an elementary particle) enters a region with a speed of 4.85 × 10^6 m/s and then is slowed at the rate of 1.63 × 10^14 m/s^2. How far does the muon take to stop?

2 Answers
Sep 7, 2015

#0.0721"m"#

or #72.1"mm"#

Explanation:

#v^2=u^2+2as#

#s=(v^2-u^2)/(2a)#

#s=(0-(4.85xx10^6)^2)/(2xx(-1.63xx10^(14)))#

#s=7.21xx10^(-2)"m"#

#s=72.1"mm"#

Sep 8, 2015

The muon take .0721 m to stop

Explanation:

#v_o# = 4.85 x #10^6# m/s

        I-------------------------------------------------------------------------l

#x_o# = 0 ------------------------------------------------------------------ v=0

a = 1.63 x #10^14# #m/s^2# ---------------------------------------------------- x=?

Don't worry about the dashes between a and x
and #x_o# and v

#x_o# is initial position
#v_o# is initial velocity
v is the final velocity when the muon stop
x is the position muon stop (in meters)
a is the acceleration

To find out how far does the muon need to stop, we could use the constant acceleration equation #v^2# = #(v_o)^2# + 2a(x - #x_o#)

Why this equation, since we know what v, #v_o#, a, #x_o# is, we can use them to solve for x.

If we plug in v and #x_o#, the equation will look like 0 = #(v_o)^2# + 2a(x-0)

We can rewrite the equation as x = (#(-(v_o))^2 / (2a)#)

so
x = #-(4.85 * 10^6)^2 / (2(-1.63 * 10^14) #

So x = .0721 m

This is just one way you could solve the answer.