A electron moving along the x axis has a position given by x = 20 te-1.4 t m, where t is in seconds. How far i?

A electron moving along the x axis has a position given by x = 20 te-1.4 t m, where t is in seconds. How far is the electron from the origin when it momentarily stops?

Update:

X = 20 te^-1.4t

Relevance

Would that be x = 20t*e^(-1.4t)?

Set the derivative = 0:

dx/dt = 20*e^(-1.4t) + 20t*-1.4*e^(-1.4t) = 0

Then e^(-1.4t) - 1.4te^(-1.4t) = 0

thus 1.4t = 1 ==> t = 5/7

I get 5.25542067766274 m at t = 5/7 s.

• cepero
Lv 4
3 years ago

a) the rate is dx(t)/dt = x'(t), so which you would be able to differentiate x(t). b) resolve the equation x'(t) < 0. this provide you with a situation on x. c) Differentiate x'(t) to get x''(t) and set x''(t) = 0. resolve and take any answer the place x'(t) <> 0. Notes: a million) the region is a 4th-order function with 4 roots a million, a million, a million,a million.5, 2) the rate is a third-order function with 3 roots a million, a million and x 3) The acceleration is a 2d-order function with 2 roots a million, y