physics problem?

Question

The electrons that produce the picture in a
TV set are accelerated by a very large electric
force as they pass through a small region in
the neck of the picture tube. This region is
1.7 cm in length, and the electrons enter with
a speed of 100000 m/s and leave with a speed
of 1.5 * 10^8 m/s.
What is their acceleration over this 1.7 cm
length? Answer in units of m/s2.

part 2
How long is the electron in the accelerating
region? Answer in units of s.

Answer 1

velocity ² = (original velocity)² + 2(acceleration)*(change in distance)

(v²-vo²)/2x = a
([1.5*10ˆ8]² - [100,000]²)/2(.017)=a
2.25*10ˆ17-1.0*10ˆ11/.034=a
a=6.6176*10ˆ18 m/s²

t=(v-vo)/(average acceleration)=(1.5*10ˆ8-100000)/(6.6176*10ˆ18)

t = 2.26517*10ˆ-11 s

Answer 2

Thus if final speed V=0.5*c, c=3*10^8 m/s is speed of light, we could somehow consider it as non-relativistic case! Then L = u*t + a*t^2/2, where given u=1*10^5m/s is initial speed of electron, given L=(1.7/100)m is its path in the accelerating gorge, a is acceleration, t is travel time through the gorge, while v=u + a*t = 1.5*10^8 is given final speed; hence t=(v-u)/a and L=u*(v-u)/a + a*(v-u)^2 / (2*a^2) = (v-u)(u + v/2 –u/2)/a = (v^2 – u^2) / a, hence a = (v^2-u^2)/L = 1.32*10^18 m/s^2, t=L/(v+u) = 0.88*10^(-10) s; I’m poor with numbers; check it!

Answer 3

find change in kinetic energy
which is equal to work done
now, work= force x displacement
displac=1.7 cm
calculate force
now, acceleratio= force/mass
(take care of units)

part 2

Answer 4

Vf^2 = Vi^2 + 2ad
(Vf^2 - Vi^2)/2d = a
((1.5*10^8m/s)^2 - (10000m/s)^2)/2(.017m) = a
a = 6.61764412*10^17m/s^2

d=.5(Vf + vi)t
(2d)/(Vf + Vi) = t
(2*.017m)/(1.5*10^8m/s + 10000m/s) = t
t =2.26515656*10^-10s