An electron in a cathode-ray-tube (crt) accelerates uniformly from 8.00 ✕ 104 m/s to 6.00 ✕ 106 m/s over 3.00 cm. (a) in what time interval does the electron travel this 3.00 cm?

Acceleration is the rate of change in velocity in respect to time. It is calculated by getting the difference in final velocity and the initial velocity and dividing it with the time taken. In this case, we can use the equations for linear motion
S = 1/2 ( U + V) t, where s is the displacement, u is the initial velocity, v is the final velocity an t is the time taken.
0.03 m = 1/2 ( 80000 + 6000000) t
0.03 m = 3040000t
        t = 9.868 ×10^-9 seconds

Initial velocity u = 8*10^4 m/s. Final velocity v = 6*10^6 m/s. Given distance s = 3 cm = 0.03 m. Let time taken be t. We use the following kinematic equation where time is the only unknown: s = 0.5*(u + v)*t. Substituting values, we obtain t as: t = 2*s/(u +v) = 0.06/(608*10^4) t = 9.8*10^-9 seconds or 9.8 nano-seconds (ns)

The answer is  a.first and third quarters