Question

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?

Answer 1

`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"`

Answer 2

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.