electrons are accelerated from rest over a distance of 800m to a velocity of 3x10 7 m/s. what is the acceleration of the electrons and what formula did you use to find the acceleration.
(1/2)[V(final)^2 - V(initial)^2] = a X
V(final) = sqrt (2 a X)
a = [V(final)^2/(2 X)]
=5.63*10^11 m/s^2
thank you, I was trying to start with finding the time first and couldn't figure it out. I tried changing the average velocity formula to come up with t. t=change in displacement / velocity, but wasn't sure if that was a proper way to change the formula, definitely didn't come up with the same answer. thanks.
Your method should have also worked. The distance covered is Vaverage = Vfinal/2 = a T/2.
Vaverage* T = X
Now eliminate T
Vaverage = (a/2)*(X/Vaverage)
Vaverage^2 = (a/2) X
Vfinal^2 = 2 a X
To find the acceleration of the electrons, we can use the equation V(final)^2 = V(initial)^2 + 2aΔX, where V(final) is the final velocity, V(initial) is the initial velocity (which is zero in this case since the electrons start from rest), a is the acceleration, and ΔX is the change in distance.
Rearranging the equation, we have 2aΔX = V(final)^2 - V(initial)^2. Since the initial velocity is zero, the equation simplifies to 2aΔX = V(final)^2.
Substituting the given values into the equation, we have 2a(800m) = (3x10^7 m/s)^2.
Rearranging the equation to solve for a, we get a = (V(final)^2) / (2ΔX).
Plugging in the values, we get a = (3x10^7 m/s)^2 / (2*800m) = 5.63x10^11 m/s^2.
The formula used to find the acceleration is V(final)^2 = V(initial)^2 + 2aΔX.