Question

An electron in a cathode-ray-tube (CRT) accelerates uniformly from 2.00 ? 10^4 m/s to 6.00 ? 10^6 m/s over 1.00 cm. A) In what time interval does the electron travel this 1.00 cm? B) What is its acceleration?

Answer 1

a & b. To find the time, I have to first solve for acceleration, using this formula:

v^2 = vi^2 + 2ad


initial velocity (vi) = 2 x 10^4 m/s

final velocity (v) = 6 x 10^6 m/s

distance (d) = .01 meters


Now re-arrange the formula to solve for "a" (acceleration).

(v^2 - vi^2) / 2d = a

(6x10^6)^2 - (2x10^4)^2 / 2 x .01 = 3.59 x 10^13 m^2/s (ACCELERATION)


Now that I have acceleration, I will use this formula to solve for time:

v = vi + at

t = (v - vi) / a

t = (6x10^6 - 2x10^4) / 3.59 x 10^13 = 1.66 x 10^-7 seconds (TIME))