# An electron moves along the x axis and a certain position equation, how far is the electron from the origin...?

an electron moves along the x axis has a position given by x= 16te^-t m, where t is time in seconds. How far is the electron from the origin when it momentarily stops?

### 1 Answer

- Anonymous9 years agoFavorite Answer
I only saw your question late friday and could not post reply until today.

x = 16te^-t

The velocity is the derivative of x with respect to (wrt) t: v = dx/dt.

This will be zero when the particle is at rest.

First find dx/dt

x is a product of two functions: 16t and e^-t. So, we need to use the product rule of differentiation which says that:- dy/dx = udv/dx + v du/dx or, in our case, since we are dealing with x not y and t not x then this is modified to:-

dx/dt = udv/dt + v du/dt ....(1)

let: u = 16t and v = e^-t; then du/dt = 16 and dv/dt = -e^-t

substituting into (1) gives us:-

dx/dt = [16t (-e^-t)] + [e^-t x 16]

dx/dt = -16te^-t + 16e^-t

dx/dt = 16e^-t(1-t)

Now, dx/dt =v and this must be zero when the particle is at rest.

Therefore,

0 = 16e^-t(1-t)

This occurs when the brackets term is zero since: 16e^-t x (0) will then = 0

1-t =0

-t = -1

t =1

The particle is stationary at t = 1 second

You can test this result by putting in values for t into x = 16te^-t and evaluating x when t is just below and above 1 sec:-

t = 0.8; x = 5.75; increasing

t = 0.9; x = 5.85; increasing

t = 1.0; x = 5.89; stopped

t = 1.1; x = 5.85; decreasing

t = 1.2; x = 5.78 ; decreasing

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