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How does one find speed and time when using two different speeds during a time interval?
In a cathode ray tube, electrons are accelerated from rest by a constant electric force of magnitude 6.4 multiplied by 10-17 N during the first 2.2 cm of the tube's length; then they move at essentially constant velocity another 43 cm before hitting the screen.
(a) Find the speed of the electrons when they hit the screen.
1 m/s
(b) How long does it take them to travel the length of the tube?
2 ns
"Panic mode"'s method is correct, but here's an alternative way that doesn't require knowing the charge on the electron nor calculating the acceleration.
Use the work-energy theorem:
Change in object's KE = work done on object
(The "object" is the electron.)
Change in KE = (final KE) − (initial KE)
= ½m(v_final)² − 0
= ½m(v_final)²
Work done = Force × (distance over which force acts)
= F×d
(where F=6.4×10^-17 N, and d = 2.2 cm = 0.022 meters)
So:
½m(v_final)² = F×d
Solve for "v_final", and plug in the given values of "F" and "d", and look up the value for "m" (mass of an electron) and plug that in. That takes care of part "a"
For part "b": Add up the two times:
t1 = time it takes to travel the first 2.2cm
t2 = time it takes to travel the next 43cm
For t1, use the formula:
time = distance / (average speed)
In cases where the acceleration is CONSTANT (as it is in this case, because the force is constant), we have:
average speed = (v_initial + v_final) / 2
So:
t1 = d × 2 / (v_initial + v_final)
or (since v_initial = 0),
t1 = d × 2 / v_final
Plug in 0.022 meters for "d", and plug in the "v_final" that you found above.
For "t2", use this formula:
time = distance / (average speed)
in this case, the distance is 43cm, and the speed in constantly v_final, so that's also the average speed. SO:
t2 = (0.43meters) / v_final
(Use the same "v_final" as before.)
The answer to "(b)" is the sum of t1 and t2.
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