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## Homework Statement

consider an electron traveling away from the origin along the x-axis in the xy plane with initial velocity vi= viî. As it passes through the region x = 0 to x = d, the electron experiences acceleration a = axî + ayj, where ax and ay are constants. For the case vi = 1.80 ✕ 10^7 m/s, ax = 8.00✕ 10^14 m/s2, and ay = 1.60 ✕ 10^15 m/s2, determine the following, at x = d = 0.0100 m. (a) the position of the electron, (b) the velocity of the electron, (c) the speed of the electron, and (d) the direction of travel of the electron (i.e., the angle between its velocity and the x axis)

## Homework Equations

Xf = Xi + Vt + 1/2at2

Vf2= vi2 + 2ax

Vf= Vi + at

## The Attempt at a Solution

Vf2 = vi2 +2ax

Vf = sqrt((1.80x10^7)^2 +2(8.00x10^14)(0.01))

Vf = 18×10^6Time taken to reach this velocity:

Vf = vi + at

t = 225s

Therefore y position after 225s

Yf = yi + viyt +1/2at2

yf = 0 + 1/2(1.60×10^5)(225)^2

Yf = 4.05× 10^9

rf = (0.01î + 4.05x10^9j)m

Magnitude = 4.05×10^9m

(b) vfx = 18x10^6î

Vfy= viy +at

Vfy = 0 + (1.60×10^5)(225)

Vfy = 36×10^6j

Vf = (18× 10^6)î + (36x10^6)j

Magnitude = 40.25 ×10^6

Direction tan@ = 36x10^6/18×10^6

@= 63.43 NE

Are these answers right or did I make a mistake