An electron in a cathode-ray tube is traveling horizontally at 2.10 x 109 cm/s when deflection plates give it an upward acceleration of 6.39 x 1017 cm/s2
How long does it take for the electron to travel 6.20 m?
What is the vertical displacement?
I got the time to be 2.95*10^-9, but I don't understand how to get the vertical displacement.
use this formula...y = 1/2 at^2
Solution please
To find the time it takes for the electron to travel 6.20 m, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = displacement (6.20 m)
u = initial velocity (2.10 x 10^9 cm/s converted to m/s)
a = acceleration (6.39 x 10^17 cm/s^2 converted to m/s^2)
t = time
Rearranging the equation:
t^2 + (2u/a)t - (2s/a) = 0
Applying the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Substituting the values:
t = [-(2u/a) ± √((2u/a)^2 - 4*(-2s/a))] / (2*(-2s/a))
Simplifying:
t = [-(2u/a) ± √((2u/a)^2 + 8s/a)] / (2*(-2s/a))
t = (u/a) ± √((u/a)^2 + 4s/a)
t = (2.10 x 10^9 cm/s converted to m/s) / (6.39 x 10^17 cm/s^2 converted to m/s^2) ± √(((2.10 x 10^9 cm/s converted to m/s) / (6.39 x 10^17 cm/s^2 converted to m/s^2))^2 + 4(6.20 m) / (6.39 x 10^17 cm/s^2 converted to m/s^2)
Evaluating the above expression will give us the time it takes for the electron to travel 6.20 m.
To find the vertical displacement, we can use the equation of motion in the vertical direction:
s = ut + (1/2)at^2
Where:
s = vertical displacement
u = initial velocity in the vertical direction (0 since the electron starts horizontally)
a = acceleration in the vertical direction (the upward acceleration given by the deflection plate)
t = time calculated above
Substituting the values and solving the equation will give us the vertical displacement.
To find the time it takes for the electron to travel a distance of 6.20 m, you can use the formula:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Given:
Initial velocity (u) = 2.10 x 10^9 cm/s (given as horizontal velocity)
Acceleration (a) = 6.39 x 10^17 cm/s^2 (given as upward acceleration)
Distance (s) = 6.20 m = 620 cm (convert m to cm)
Rearranging the formula, we get:
(time^2) * (0.5 * a) + (u * time) - s = 0
This is a quadratic equation in terms of time. We can solve it using the quadratic formula:
time = (-b ± √(b^2 - 4ac)) / (2a)
where a = 0.5 * a, b = u, and c = -s
Plugging in the given values, we have:
a = 0.5 * 6.39 x 10^17 = 3.20 x 10^17
b = 2.10 x 10^9
c = -620
Substituting these values into the quadratic formula, we get:
time = (-2.10 x 10^9 ± √((2.10 x 10^9)^2 - 4 * (0.5 * 3.20 x 10^17) * (-620))) / (2 * (0.5 * 3.20 x 10^17))
Simplifying the equation further, we get:
time = (-2.10 x 10^9 ± √(4.41 x 10^18 - 3.20 x 10^17 * (-620))) / (3.20 x 10^17)
time = (-2.10 x 10^9 ± √(4.41 x 10^18 + 1.984 x 10^20)) / (3.20 x 10^17)
time = (-2.10 x 10^9 ± √(2.4284 x 10^20)) / (3.20 x 10^17)
Since the electron is traveling horizontally, we will take the positive root. Taking the square root and simplifying, we get:
time = (-2.10 x 10^9 + 4.928 x 10^10) / (3.20 x 10^17)
time = 2.905 x 10^-9 seconds (approx.)
So, the time it takes for the electron to travel a distance of 6.20 m is approximately 2.905 x 10^-9 seconds.
Now, let's calculate the vertical displacement of the electron.
Vertical displacement = (1/2) * acceleration * time^2
Substituting the values we know:
Vertical displacement = (0.5 * 6.39 x 10^17) * (2.905 x 10^-9)^2
Vertical displacement = (3.195 x 10^17) * (8.433 x 10^-18)
Vertical displacement = 2.688 x 10^-1 cm
Therefore, the vertical displacement of the electron is approximately 0.2688 cm.