Horizontal Kinematics - Magnitude of acceleration?

Question

An electron, starting from rest and moving

with a constant acceleration, travels 5.1 cm in

2.2 ms.

What is the magnitude of this acceleration

in km/s2? Answer in units of km/s2.

Answer 1

The final velocity of the electron is:

v = 0.051 m / 0.0022 s = 23 m/s

The acceleration of the electron is:

a = (vf² - vi²) / 2d = (23 m/s)² / (2 x 0.051 m) = 5,200 m/s²

5,200 m/s² / 1,000 = 5.2 km/s²

Answer 2

using kinematics equation

S=ut + 0.5 at^2

5.1= 0 + 0.5 a * (2.2)^2

This would give a in cm/(ms)^2

1ms=1/1000 s

and 1cm= 1/100000 Km

so a= {(10.2)/(2.2)^2} 10 KM/s^2

Answer 3

we can use the equation:

dist = vot+1/2 at^2

v0=0, a is to be found, dist = 5.1 cm=0.051m, t=2.2x10^-3 s

so we have

a=2d/t^2 =

2(0.051)/(2.2x10^-3)^2

=21,074 m/s/s

=21.1 km/s/s