What is the acceleration of an electron in a cathode ray tube?

[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 into give you acceleration?

 

[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?

 

[runner1738]

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

 

[Doc Al]

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

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?

 

[Curious3141]

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

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.

 

[christinono]

Actually, the best formula to use would be:

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

This way, only 2 step would be required.

 

[Curious3141]

Actually, the best formula to use would be:

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

This way, only 2 step would be required.

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

 

[christinono]

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

That's right. Sorry about that...

 

[runner1738]

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

 

[christinono]

How about this formula:
$$V_f = V_i + at$$