Find acceleration given initial and final velocity, and displacement

[donking225]

1. An electron in the cathode ray tube of a television set enters a region where it accelerates uniformly from a speed of 62300 m/s to a speed of 3.86 × 10⁶ m/s in a distance of 3.17 cm.
What is its acceleration?

2. a = [Vf² - Vi²] / [2d]



3. a = [(3.86 × 10⁶)m/s² - 62300m/s²] / [2 * 0.0317m]
I'm getting the answer 2.35(rounded to hundredth place)*10¹⁴m/s² , but it doesn't seem to be correct when I put that answer on my homework. Please help.

 

[siddharth23]

Acceleration is rate of change of velocity. It's (vf-vi)/t . Don't square the velocities. And don't divide by the distance.

More applicable formula here would be
v^2=u^2 + 2as (s is the distance)

 

[donking225]

Acceleration is rate of change of velocity. It's (vf-vi)/t . Don't square the velocities. And don't divide by the distance.

More applicable formula here would be
v^2=u^2 + 2as (s is the distance)

Wouldn't you still divide by (2s) anyways based on algebra. To get (a) alone you would subtract u^2 from both sides of the equation and have v^2-u^2= 2as. Then you would divide both sides by (2s) to get (a) alone. Isn't that basically the same as the original equation I had? I'm also confused about what you mean by not squaring the velocities.

 

[ehild]

1. An electron in the cathode ray tube of a television set enters a region where it accelerates uniformly from a speed of 62300 m/s to a speed of 3.86 × 10⁶ m/s in a distance of 3.17 cm.
What is its acceleration?

2. a = [Vf² - Vi²] / [2d]



3. a = [(3.86 × 10⁶)m/s² - 62300m/s²] / [2 * 0.0317m]
I'm getting the answer 2.35(rounded to hundredth place)*10¹⁴m/s² , but it doesn't seem to be correct when I put that answer on my homework. Please help.

The result is correct.

ehild

 

[donking225]

The result is correct.

ehild

That's what I thought too, I guess the only explanation to why I'm getting this wrong on my homework website is that I am entering it in an incorrect format.