Acceleration given velocity and displacement

[sjoseph08]

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Good Afternoon,

I am getting two different results when completing this question and I don't understand why both methods don't get to the same answer.

An electron with an initial velocity of 150,000 m/s enters a region of length L=.01 m where it is electrically accelerated. It emerges with a velocity of 5,700,000 m/s. What is its acceleration, assumed constant?

So the method I took was to take the displacement equals average velocity multiplied by time. I solved for time and got 0.000000003 s. I then did change in velocity divided by time to get the acceleration. (5.5X10^6/3.0X10^-9)= 1.8X10^15 m/s^2. I checked the answer and it's incorrect.

So I went back and read the test which suggested using displacement equals velocity final squared minus velocity initial squared divided by 2a. I solved for a and got velocity final squared minus velocity initial squared divided by 2 times the displacement. That result is 1.62 X 10^15, which is the correct answer per the book.

Oddly, I get 3X10^-9 as the time regardless of which answer I use. Why would it matter which approach I used if acceleration is assumed constant?

Thank you for your help!

 

[milski]

You have rounded the time too much in the first solution. You can show algebraically that both ways are equivalent. If you want to work with number, using a few more significant digits will get you the same result.

If you look at it, you're using only one sig digit for time, so both of your results are the same up to a single significant digit.

A more precise number for the time is: t=3.4188034x10^-9

 

[sjoseph08]

Thank you very much. I just couldn't figure out the discrepancy.