Physics - kinematics homework question.

[L_0611]

Homework Statement

An electron is accelerated from rest to a velocity of 2.0x10^7 m/s.
a) if the electron traveled 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=?

Homework 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.

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

 

[justsomeguy]

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

 

[SteamKing]

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

 

[L_0611]

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.

 

[justsomeguy]

This is very wrong.

4x10^16m/s = 0m/s+2a(0.10m)
4x10^16m/s / 0.10m = 2a
4x10^18m/s /2 = a

In v2 = v1 + 2ad, solving for a, you need to do all the steps right and in the right order:

v2 = v1 + 2ad
v2-v1 = 2ad
(v2-v1)/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 non-rest velocity.

 

[L_0611]

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

(V2-V1)/2d=a
(2x10^7m/s-0m/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/s-0m/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.

 

[justsomeguy]

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?

 

[L_0611]

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.

 

[justsomeguy]

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.

 

[L_0611]

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...?

 

[justsomeguy]

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.

 

[L_0611]

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.

 

[haruspex]

(2x10^7m/s)²=(0m/s)²+2a(0.10m)
4x10^16m/s = 0m/s+2a(0.10m)

How do you get 10¹⁶?

4x10^16m/s / 0.10m = 2a
4x10^18m/s /2 = a

How do you get from 10¹⁶ to 10¹⁸?

Part b:
t=aV

Oh no it isn't. Dimensionally that makes no sense. Acceleration has dimension LT^-2. Speed has dimension LT^-1. If you multiply those together you'll have L²T^-3, not the T that you want. Is this the equation you have in your notes?

 

[L_0611]

okay so I tried it again, this time I rearranged the equation d=V1t+1/2at² to read a=(V2²-V1²)/(2d) when I plug in my numbers into this equation this is what I get

a=(V2²-V1²)/(2d)
= (2x10^7m/s-0m/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.

 

[L_0611]

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.

 

[justsomeguy]

okay so I tried it again, this time I rearranged the equation d=V1t+1/2at² to read a=(V2²-V1²)/(2d) when I plug in my numbers into this equation this is what I get

a=(V2²-V1²)/(2d)
= (2x10^7m/s-0m/s)/(2x0.10m)
= (2x10^7m/s)/(0.20m)
= 1x10^8m/s²

This now looks correct.

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.

Look another post up. To be short and sweet, this is just plain wrong. You've been asked to find out how long it took the electron to traverse the 0.1m distance during which it was accelerated. How do you figure the answer to this is time = acceleration * velocity?

 

[SteamKing]

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?

 

[L_0611]

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.

 

[justsomeguy]

Oh well if you forgot to do something, go back and.. do it, sorry. I just meant the arithmetic looked right. As I tried to explain before, I didn't bother typing out the powers since I (irresponsibly) just see them and assume that operation is done before starting to simplify.

 

[L_0611]

ok perfect, I re did it using the same formula and squaring both V1 and V2. Thanks for your help, I really appreciate it.

 

[haruspex]

i did use the formula t=aV,

Do you now understand that formula is completely wrong? Have you corrected it in your notes?

 

[L_0611]

wait what? I thought the original formula is a=V/t and if I rearrange it to solve for t wouldn't I end up with time=acceleration x velocity. If that is wrong like you are saying haruspex, can you please explain why, thank you.

 

[justsomeguy]

Can you show that work? How did you rearrange a=V/t to arrive at t=aV?

 

[L_0611]

wait a second I just redid that and realized it should be t=V/a not t=aV. Thanks for making me realize my mistake.

 

[justsomeguy]

wait a second I just redid that and realized it should be t=V/a not t=aV. Thanks for making me realize my mistake.

It is really a=dV/dT (a = Δv/Δd) but since you're starting v and d both from 0 it doesn't matter in this case. That's why I brought it up last page -- but you said it wasn't on your worksheet/in your notes. ;)

So... what's your answer now?

 

[L_0611]

so are you saying that time should equal change in velocity divided by change in distance, not acceleration? but because both my initial velocity and initial distance is 0, I can just use time= velocity/acceleration? I think I'm somewhat starting to understand it, however still very confused.

 

[justsomeguy]

so are you saying that time should equal change in velocity divided by change in distance, not acceleration? but because both my initial velocity and initial distance is 0, I can just use time= velocity/acceleration? I think I'm somewhat starting to understand it, however still very confused.

You should start at what the definitions of the terms are. Velocity is a change in distance over time. Acceleration is change in velocity over time. Time and change in time (t and dt or Δt) for acceleration are interchangeable since you're talking about an interval. Saying it took 5 seconds, or that it took from the 5 second to the 10 second mark are the same thing.

(I am going to NOT leave out the squares for you here. If they are not there, they are not meant to be.)

Acceleration is defined as a = Δv/Δt and in your problem is meters and seconds, so acceleration will be in meters per second per second, which is the same as m/s/s, which is also the same as m/s^2.

Velocity is v2 = v1 + at, in other words, final velocity is equal to starting velocity plus the rate of change of velocity multiplied by how long you were moving. If you look at this in relation to the bit about acceleration, what do you see?

Distance is defined one of two ways. For constant velocity it's simply d = vt. How far you go is your velocity times how long you're moving. For constant acceleration (changing velocity), the formula is d = 1/2(v2-v1)t.

If it's easier for you (and it is for me), what that really is saying is it's average velocity times time. Rearrange the terms and you'll see that 1/2(v2-v1)t is the same as ((v2-v1)/2)t.

Those both solve for distance when you know time and the start and end velocities.

When you know acceleration and velocity but not the distance, you use the formula you initially tried to use: v2² = v1² - 2ad.

If you practice with all of these and make sure that now matter how you do the substitutions, the answers always come out the same. As long as you follow the rules, they always will, because all of these equations are derived from one another.

 

[L_0611]

ok perfect, thanks for the explanation, I think I get it now. I really appreciate you taking your time to help me out with this question and obviously I was right out to lunch, I am taking this course through correspondence so I don`t have the privilege of asking a teacher, so I really appreciate your help. Thanks again.

 

[haruspex]

so are you saying that time should equal change in velocity divided by change in distance, not acceleration?

No, that's dimensionally wrong too (it would give 1/time) and even if you divided the other way it would be wrong.
If distance s is a function of time, t, s = s(t), the general equations are:
v(t) = ds/dt
a(t) = dv/dt = d²s/dt².
In the case where a is constant:
v is a linear function of t: Δv = a Δt; v(t) = v(0) + a t
s is a quadratic function of t: s(t) = s(0) + v(0)t + a t²/2.

 

[L_0611]

so I just did my homework question and this are the answers I got:

Part a:
V2²=V1²+2ad
a=(V2²-V1²)/2d
a=(2.0x10^7²m/s-0m/s²)/2(0.10m)
a=(4x10^14m/s)/0.20m
a= 2x10^15m/s²
therefore the electron's acceleration is 2x10^15m/s²

Part b:

a=ΔV/Δt
Δt=ΔV/a
Δt=(2.0x10^7m/s-0m/s)/(2x10^15m/s²)
Δt=(2.0x10^7m/s)/(2x10^15m/s²)
Δt=0.00000001s

therefore it took the electron 0.00000001s to attain its final velocity.

Everyone that has commented on here has been of great help and your time, effort and help is greatly appreciated. Thank you, please correct me if these are not the correct answers for my problem, however from what I've gather from you on here, I believe these to be correct, again please correct me if I'm wrong and thank you for your time and help.

 

[justsomeguy]

That all worked out fine for me. Hopefully someone else will come along to see if there are any mistakes.