Why Is My Calculation of Electron Travel Time Incorrect?

[cheese825]

I cannot figure out what is wrong! (frustrated)

Homework Statement

An electron in a cathode-ray-tube (CRT) accelerates from 4.00 x 10^4 m/s to 6.00 x 10^6 m/s over 2.50 cm

Homework Equations

How long does the electron take to travel this 2.50 cm?

The Attempt at a Solution

(4.00 x 10^4 + 6.00 x 10^6) / 2 = avg velocity

2.50 x 10^-2 / avg velocity = answer

Why is that wrong?
(I assume the acceleration is meant to be constant)

 

[Tac-Tics]

Homework Statement

An electron in a cathode-ray-tube (CRT) accelerates from 4.00 x 10^4 m/s to 6.00 x 10^6 m/s over 2.50 cm

Homework Equations

How long does the electron take to travel this 2.50 cm?

This is the more appropriate place to put this question =-D

(I assume the acceleration is meant to be constant)

I am also making this assumption. I don't think you can do the problem without it, and I believe this is more or less true of an CRT.


So, this is how I set up the problem. You have two velocities of the electron. You can figure out the change in kinetic energy by \frac{1}{2} m (v_1^2 - v_2^2) (m is the mass of an electron). Kinetic energy put into an object is called due to a force through a distance is called work. Work is described by the equation W = Fd, (d in this case is our 2.50cm).

Using those equations, we can isolate the force on the electron. From there, the rest of the problem is pretty straightforward.

Good luck!

 

[cheese825]

Thanks Tac Tics.

But it seems you are making this more complicated than it should be. This is supposed to be simple 1 dimensional motion problem, no need for Work or even Force.

Is it not true that given 2 velocities, the initial and the final, that if you find the average of those and then multiply it by the time, you should get the total distance traveled?

I am at the point where I think I'll blame the book for publishing a wrong answer and making me lose a quarter of my brain mass via combustion.

I have gotten an answer (8.2781 * (10^-9) and am convinced it is correct after working through it again and again. I'll gladly give sexual favors to whoever gets me the correct answer.

 

[loonychune]

s = \frac{1}{2}(u+v)t

...one of the SUVAT equations for constant acceleration... this is esentially what you've used to find t... and it does give 8.278 * 10^{-9}... so i'd probably move on to something else unless someone else has owt to say

 

[Tac-Tics]

Thanks Tac Tics.

But it seems you are making this more complicated than it should be. This is supposed to be simple 1 dimensional motion problem, no need for Work or even Force.

I should have disclaimed my answer would not be efficient (I'm no physics major).

But at the least, it should give you something to double check yourself on.

Although, one technical point, you don't need to *know* the mass of an electron. You have
\frac{1}{2}m(v_1^2 - v_0^2) = W = Fd = mad (hehe, "mad"), but you can cancel the m's to get \frac{1}{2}(v_1^2 - v_0^2)= ad, isolating acceleration that way.

Then the distance at a time measured from the first point is given by x(t) = \frac{1}{2} at^2 + v_0t = \frac{1}{4d}(v_1^2 - v_0^2) t^2 + v_0 t. So unless I've made a mistake, plugging into your calculator: x(t) where t is your final answer should give you d. This will help you confirm if you honestly think you found an error in your book.

I'm trying this, and I'm not coming up with the right answer either. I will look at it more closely when I get home.

 

[Tac-Tics]

The Attempt at a Solution

(4.00 x 10^4 + 6.00 x 10^6) / 2 = avg velocity

2.50 x 10^-2 / avg velocity = answer

I double-checked the algebra, and this is correct.

From my above post, if you plug in d / v_avg into the function x(t), you get back d, so after d/v_avg seconds.