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Physics  kinematics homework question. 
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#1
Dec1312, 11:11 PM

P: 24

1. The problem statement, all variables and given/known data
An electron is accelerated from rest to a velocity of 2.0x10^7 m/s. a) if the electron travelled 0.10m while it was being accelerated, what was its acceleration? b)How long did the electron take to attain its final velocity? V1= 0m/s V2= 2.0x10^7m/s d= 0.10m a=? t=? 2. Relevant equations Please correct me if I'm wrong but I believe the equations I have to use are V2²=V1²+2ad for part a and V=d/t for part b. Thanks for your help. 3. The attempt at a solution I attempted this questions and obtained the following answers, but not sure if my formulas are correct. Part a) 2x10^18 m/s² Part b) 4x10^27 s 


#2
Dec1312, 11:17 PM

P: 166

Can you 'show your work' in plugging in the values and working through them to arrive at the answers you have?



#3
Dec1412, 12:14 AM

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You might want to check part b. It's longer than the age of the universe by many orders of magnitude.



#4
Dec1412, 01:15 PM

P: 24

Physics  kinematics homework question.
when I tried this problem this is how I arrived to my answers (showing my work)
Part a: V2²=V1²+2ad (2x10^7m/s)²=(0m/s)²+2a(0.10m) 4x10^16m/s = 0m/s+2a(0.10m) 4x10^16m/s / 0.10m = 2a 4x10^18m/s /2 = a 2x10^18m/s² = a Therefore the acceleration is 2x10^18m/s² Part b: t=aV = 2x10^18m/s² (2x10^7m/s  0m/s) = 2x10^18m/s² (2x10^7m/s) = 4x10^27s therefore it took the electron 4x10^27s to reach its final velocity. SteamKing, what do you mean when the age is longer than the age of the universe by many orders of magnitude? do you mean that my answer should be a bigger number or that it is too big of a number? thanks for your help. 


#5
Dec1412, 01:39 PM

P: 166

This is very wrong.
v2 = v1 + 2ad v2v1 = 2ad (v2v1)/2d = a Edit: I'm not sure why you don't approach it as simply a = dV/dT however. The formula you're using is more intended for something starting with a nonrest velocity. 


#6
Dec1412, 02:03 PM

P: 24

I tried doing it the way you said and got the same result if I square both velocities which is what my original formula says, if I don't square them like you wrote on the last post this is the answer I get
(V2V1)/2d=a (2x10^7m/s0m/s) / 2(0.10m) = a (2x10^7m/s²)/0.20m = a 1x10^9 m/s² = a as for part b if I use the new result i got for acceleration the answer I get would be t=aV =(1x10^9m/s²)(2x10^7m/s0m/s) = (1x10^9m/s²)(2x10^7m/s) = 2x10^18s as for the other formula you suggested, I do not have that formula on the formula sheet I was given with my work and by looking at it I can see that I time is one of the variables, therefore I would have 2 unknown variables if I use that formula and not sure how to approach it that way, please explain if you know how. Thank you. 


#7
Dec1412, 02:16 PM

P: 166

That wasn't to indicate you shouldn't square them, I was just assuming you'd do that right away since you know both values. I won't suggest you use something not on your worksheet, instructors tend to frown on that.
Sorry for perhaps leading you down the wrong path. Is d = vt+1/2at^2 on your worksheet? 


#8
Dec1412, 03:26 PM

P: 24

yes I have d=vt+1/2at^2 on my formula sheet, however I didn't try using it because I figured since I don't yet have a value for t I would end up with two unknown variables, I like I said before I'm not sure how to approach that. Sorry, I'm just really confused with this question, I graduated high school a while ago and trying to apply for an employment opportunity and they require me to take a physics course (which I never took in school) and upgrade my math, that is why I am struggling so much with this. I'd like to thank you for your time, effort and help, I really appreciate it.



#9
Dec1412, 03:30 PM

P: 166

The idea is to solve for one of the variables (a or t) and then substitute the answer into the 2nd equation in its place. Solve for that, then use the value you get in the first equation to solve it as well.



#10
Dec1412, 03:36 PM

P: 24

I'm not exactly sure how I would just solve for one variable when I need the other variable for the answer. Am I completely out to lunch and missing something here...?



#11
Dec1412, 03:39 PM

P: 166

If you start with x=yz and you only know x and are trying to find y, rearrange it to y=x/z. In the second formula, say it's z=2y+x, you substitute in your answer from the first one, so z=2y+x becomes z = 2(x/z) + x.
Calculate z and then plug the answer back into y=x/z. 


#12
Dec1412, 05:16 PM

P: 24

I'm sorry but I'm still not really getting it, I understand what you're saying about the two unknown variables, however I've been trying to do it with my question and still not able to figure it out.



#13
Dec1412, 06:05 PM

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#14
Dec1412, 06:08 PM

P: 24

okay so I tried it again, this time I rearranged the equation d=V1t+1/2at² to read a=(V2²V1²)/(2d) when I plug in my numbers in to this equation this is what I get
a=(V2²V1²)/(2d) = (2x10^7m/s0m/s)/(2x0.10m) = (2x10^7m/s)/(0.20m) = 1x10^8m/s² then I use that answer for part b using the formula t=aV and when I plug in my numbers this is what I get t=aV =(2x10^8)(2x10^7) =2x10^15 please let me know if this is correct, if not I'm all out of ideas. Thank you. 


#15
Dec1412, 06:17 PM

P: 24

correction to my last post, I forgot to square V2 and V1 in the first formula therefore giving me a wrong answer, I redid the equation using the same formula and values and this is the answer I got for part a, 2x10^15 and then used that for part b and obtained this answer 4x10^22.



#16
Dec1412, 06:40 PM

P: 166




#17
Dec1412, 06:43 PM

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If the time calculated for the electron to travel 0.1 m is longer than the age of the universe, that means it could not happen.
Step back for a minute and consider what the numbers mean. The electron will start from rest and reach a velocity of 20000 km / sec. Orbital velocity about the earth is only about 8 km / sec. The distance the electron has traveled by the time it reaches its final velocity is about the width of you palm. Therefore, is it reasonable to conclude that the time in which it takes the electron to travel this distance is equivalent to many billions of years? 


#18
Dec1412, 06:45 PM

P: 24

when you say "this now looks correct" are you referring to the answer or to the formula? because on the post that you mention that I forgot to square V1 and V2 therefore that is why I doubt that is the right answer hence why I'm asking if you're just referring to the formula, as for part b i did use the formula t=aV, however if my answer for part a is wrong my answer for part b will be wrong as well.



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