A muon (an elementary particle) enters a region with a speed of 5.00\times 10^6 m/s and then is slowed at the rate of 1.25 \times 10^{14} m/s^2. (a) How far does the muon take to stop?

#### Solution:

### To solve this question, we will use the following equation:

### v^2=v_0^2+2a(x-x_0) where v is final velocity, v_0 is initial velocity, a is acceleration, x is final position, and x_0 is initial position.

### We know that velocity finial is 0, and initial position is 0. Thus, we can write:

### x=-\frac{1}{2}\frac{v_0^2}{a}

### x=-\frac{1}{2}\frac{(5.99\times 10^2)^2}{-1.25\times 10^{14}}

### x=0.100 m.

###### This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 26.