physics - kinematics homework question.

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

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

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

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

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.