Physics - kinematics homework question.

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  • #26
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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.
 
  • #27
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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.
 
  • #28
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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.
 
  • #29
haruspex
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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.
 
  • #30
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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.
 
  • #31
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That all worked out fine for me. Hopefully someone else will come along to see if there are any mistakes.
 
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