An electron moving along the x axis has a position given by x = 14 te-t m, where t is in seconds. How far is the electron from the origin when it momentarily stops?

To find the distance of the electron from the origin when it momentarily stops, we need to determine the time at which the electron's velocity becomes zero. This will occur when the derivative of its position with respect to time is equal to zero.

Let's find the velocity of the electron by taking the derivative of its position equation with respect to time:

v = dx/dt = 14e^(-t) * (-1) = -14e^(-t) m/s

Now, we can set the velocity equal to zero:

0 = -14e^(-t)

To solve for t, let's divide both sides of the equation by -14:

0 = e^(-t)

Since any non-zero number raised to the power of 0 is equal to 1, we can conclude that:

e^(-t) = 1

To solve for t, we take the natural logarithm (ln) of both sides:

ln(e^(-t)) = ln(1)

Using the property that ln(a^b) = b * ln(a), we get:

- t * ln(e) = 0

Since ln(e) is equal to 1, we have:

- t = 0

By multiplying both sides by -1, we obtain:

t = 0

So, the electron momentarily stops at t = 0 seconds.

To find the distance from the origin when t = 0, we substitute t = 0 into the position equation:

x = 14e^(-t) = 14e^0 = 14m

Therefore, the electron is 14 meters away from the origin when it momentarily stops.