An electron moving along the x axis has a position given by x = 14te-4t m,?
An electron moving along the x axis has a position given by x = 14te-4t m, where t is in seconds. How far is the electron from the origin when it momentarily stops?
- kuiperbelt2003Lv 78 years agoFavorite Answer
the phrase momentarily stops means you want to find when and where the velocity is zero, to do this, we differentiate x(t) and find where this is zero (in other words, find the extremum of the x function)
v = dx/dt = 14 e^-4t - 56t e^-4t =0
14 e^-14t = 56t e^-4t
the common exponential factors cancel, and we find easilty that t=1/4s
this is the time of zero velocity; substitute t=1/4 into the x(t) equation to find:
x(t=1/4) = (14/4)e^(-1)
- ascencioLv 43 years ago
right here we pass, x=3t^4 -16t^3+24t^2 a. Differentiate each and each and every area by technique of t, speed, dx/dt = 12 t^3 - 40 8 t^2 + 40 8 t --------- (a million) Acceleration, d^2x/dt^2 = 36 t^2 - ninety six t + 40 8 ----------- (2) b. right here V = 0; from 1st eqn, 0 = 12 t^3 - 40 8 t^2 + 40 8 t so t = 0 or 2 s c. at t = 0 (right here it will be both at 0 or 2 or both, yet by technique of utilising some values we may be able to be certain it will be 0) d. right here evaluate eqn 2 = 0; that provides t = 2/3 or 2. considering the fact that 2/3 < 2, evaluate the occation even as t = 2/3, then practice that value to eqn a million, to get the answer, V = 128/9 (i.e. 14.22 m/s).