Question: An electron in a cathode-ray tube accelerates from rest with a constant acceleration of 5.33 x 1012 m/s2 for 0.15 us (1 js = 10 s). The electron then drifts with constant velocity for 0.2 us. Finally, it comes to rest with an acceleration of -2.267 x 1013 m/s2. How far does the electron travel? 401x - 5.33 x 1012 m/s2 223x = -2.67 10'3 m/s2 Por = 0 012 0 * =

An electron in a cathode-ray tube accelerates from rest with a constant acceleration of 5.33 x 1012 m/s2 for 0.15 us (1 js = 10 s). The electron then drifts with constant velocity for 0.2 us. Finally, it comes to rest with an acceleration of -2.267 x 1013 m/s2. How far does the electron travel? 401x - 5.33 x 1012 m/s2 223x = -2.67 10'3 m/s2 Por = 0 012 0 * = 0 X1 X2 1, = 0.150 us 12 = 4, +0.200 13 to = 0 Ax = 6.00 cm; » = 8.00x10' m's Ar; = 1.20 cm As, 16 cm Ar = 23.2 cm Steps Answers 1. Find the displacement and final velocity for the first Ax, = 6.00 cm; y = 8.00x10ʻm/s 0.15-jis interval 2. Use this final velocity as the constant velocity to find Ar, = 16 cm the displacement while it drifts at constant velocity. 3. Use this same velocity as the initial velocity and Ar; = 1.20 cm Equation 2-15 with v = 0 to find the displacement for the third interval in which the electron slows down. 4. Ar = 23.2 cm Add the displacements found in steps 1, 2, and 3 to find the total displacement.