Electron acceleration Qs?

1. Feb 4, 2005

runner1738

An electron in the cathode ray tube of a television set enters a region where it accelerates unifromly from a speed of 40200 m/s to a speed of 3.01 x 10^6 m/s in a distance of 2.08 cm. What is the acceleration? answers in m/s

i understand how you can do 3.01 x 10^6-40200/.00208 and get m but then where does time factor in to give you acceleration?

2. Feb 4, 2005

Pseudo Statistic

$$v^{2} = u^{2} + 2as$$
$$v = 3.01 * 10^{6} ms^{-1}$$
$$u = 40200 ms^{-1}$$
$$s = 2.08cm = 2.08 * 10^{-2} m$$
Solve for a?

3. Feb 4, 2005

runner1738

for what length of time is the electron in this region where it accelerates? in s

4. Feb 4, 2005

Staff: Mentor

As Pseudo Statistic showed, you need not calculate the time to find the acceleration. But if you wish to find the acceleration by first finding the time, that's perfectly OK. Consider that $d = v_{average} t$. What's the average velocity?

5. Feb 4, 2005

Curious3141

You're not asked that, and the method posted by Pseudo is the best way.

But if you really want to do it via the time, use :$$v = u + at$$

then $$s = ut + \frac{1}{2}at^2$$

in two steps to get the answer.

6. Feb 4, 2005

christinono

Actually, the best formula to use would be:

$$V_f^2 = V_i^2 + 2ad$$

This way, only 2 step would be required.

7. Feb 4, 2005

Curious3141

This has already been explained, and in fact, only one step would be required.

8. Feb 4, 2005

christinono

That's right. Sorry about that... :tongue2:

9. Feb 4, 2005

runner1738

i solved for a, but the second part of the question is to solve for length of time that it accelerates

10. Feb 4, 2005

christinono

$$V_f = V_i + at$$