Is 1.875×10^8 m/s² the Correct Acceleration for the Electron in the TV Tube?

[kleencut]

1. An electron of mass 9.11×10^−31 kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is a distance 1.20 cm away. It reaches the grid with a speed of 4.50×10^6 m/s. The accelerating force is constant.
Find the acceleration.
Find the time to reach the grid.
Find the net force. (ignore gravity on the electron)




2. v^2=(u^2)+2*a*s



3. I plugged in the numbers to the equation v^2=(u^2)+2*a*s. to find acceleration, but the anwer (1.875*10^8) was incorrect. Is this the correct equation? The other answers should stem from the acceleration right?

 

[rl.bhat]

(4.50×10^6 m/s.)^2 = 2*a*(1.2x10^-2)
From the above equation find the value of a

 

[minifhncc]

Acceleration - Sub into equation v²=u²+2aS -- make sure you only substitute after converting to SI units.

Time - Use Equation a=(v-u)/t (Yes you need to find acceleration first)

Net force - Use equation \sumF=ma

And 1.875*10^8 for acceleration is wrong.

Hope this helps.