An electron moving along the x axis has a position given by x=16te^{-t} , where t is in seconds. How far is the electron from the origin when it momentarily stops?

#### Solution:

### To do this question, we must know the derivative e^{bx}=be^{bx} where b is a constant.

### Thus, we can write:

### v=\frac{dx}{dt}=\left(\frac{d(16t)}{dt}\right)\cdot e^{-t}+(16t)\cdot\left(\frac{de^{-t}}{dt}\right)

### At t= 1 s, this function is 0 which is the time at which the electron stops. To find where the electron stops, we plug t=1 into our original function, x=16te^{-t}, which gives us, x=5.9 m.

###### This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 14.