anonymous
  • anonymous
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Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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4n1m0s1ty
  • 4n1m0s1ty
Ok this problem is pretty much the same as the last one.
4n1m0s1ty
  • 4n1m0s1ty
The only difference is that we are dealing with velocity and acceleration, rather than position. I think you probably already know that the acceleration is the change in velocity over time. So, in this problem we kind of treat velocity like position in the last problem, and acceleration like velocity previously.
anonymous
  • anonymous
So would it start off at 2m/s?

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4n1m0s1ty
  • 4n1m0s1ty
Yes
4n1m0s1ty
  • 4n1m0s1ty
Now what is the velocity at t = 1?
4n1m0s1ty
  • 4n1m0s1ty
The acceleration was 2 m/s2 at t = 0, so after 1 second, we give a boost to the initial 2m/s velocity.
anonymous
  • anonymous
So would it be 4 since we are adding 2 to the 2m/s of t=0?
4n1m0s1ty
  • 4n1m0s1ty
yep
anonymous
  • anonymous
t=2 would be 0
4n1m0s1ty
  • 4n1m0s1ty
Nope. What's the acceleration at t = 1?
4n1m0s1ty
  • 4n1m0s1ty
It goes from 2 m/s2 to ?
anonymous
  • anonymous
1
4n1m0s1ty
  • 4n1m0s1ty
yes, so our velocity is at 4 m/s, and we just add 1 m/s after 1 s, so the velocity at t =2 is?
anonymous
  • anonymous
5
4n1m0s1ty
  • 4n1m0s1ty
yep
4n1m0s1ty
  • 4n1m0s1ty
so now for t = 3. Look at the acceleration at t =2 and add is to your velocity.
anonymous
  • anonymous
Then would t=3 be 0 since it's a line?
4n1m0s1ty
  • 4n1m0s1ty
yes the acceleration is 0.
4n1m0s1ty
  • 4n1m0s1ty
so that means the velocity doesn't change.
anonymous
  • anonymous
Right
anonymous
  • anonymous
Would the be negative numbers since the acceleration is in the negative now?
4n1m0s1ty
  • 4n1m0s1ty
so when the acceleration is negative, we are reducing the positive velocity. So at t =3, your velocity is 5 m/s, and we subtract 3 m/s now since its negative
4n1m0s1ty
  • 4n1m0s1ty
So at t = 4, v = ?
anonymous
  • anonymous
why not 2?
4n1m0s1ty
  • 4n1m0s1ty
Are you saying why not v = 2 at t = 3?
anonymous
  • anonymous
why not subtract 2?
4n1m0s1ty
  • 4n1m0s1ty
Because the acceleration is -3 right not -2?
anonymous
  • anonymous
Sorry. Was getting confused with the +2
4n1m0s1ty
  • 4n1m0s1ty
ok so what is the velocity at t = 4
anonymous
  • anonymous
v=2
4n1m0s1ty
  • 4n1m0s1ty
yep
4n1m0s1ty
  • 4n1m0s1ty
t = 5?
4n1m0s1ty
  • 4n1m0s1ty
the acceleration is back to 0 again right?
anonymous
  • anonymous
v=2
4n1m0s1ty
  • 4n1m0s1ty
yep
4n1m0s1ty
  • 4n1m0s1ty
So i think thats it. Any questions?
anonymous
  • anonymous
Okay so 0s=2 1s=4 2s=5 3s=0 4s=2 5s=2
anonymous
  • anonymous
Sorry it took so long to type that. My browser kept freezing up on me.
4n1m0s1ty
  • 4n1m0s1ty
3s = 5, the acceleration was 0, not the velocity
anonymous
  • anonymous
Okay so if we took the velocity at 3.41s how do we go about doing that?
4n1m0s1ty
  • 4n1m0s1ty
look between t = 3 and t = 4 on the velocity graph you just drew.
anonymous
  • anonymous
Looks to be about 3.50 to me...does that seem reasonable?
4n1m0s1ty
  • 4n1m0s1ty
Maybe, I'm not sure how your assignment system works. I'm not sure if it wants you to calculate it exactly or just by estimation. Did your instructor give you any equations to work with.
4n1m0s1ty
  • 4n1m0s1ty
Just for my information is this a high school or college physics course?
anonymous
  • anonymous
He didn't. It's college but this is the first time I've ever had physics.
4n1m0s1ty
  • 4n1m0s1ty
Ok well I guess its just an estimate then.
anonymous
  • anonymous
It didn't take 3.50 so I think it's wanting more exact.
4n1m0s1ty
  • 4n1m0s1ty
Ok well, then that would require calculating the slope of the graph between t = 3 and t = 4. Do you know how to do that?
4n1m0s1ty
  • 4n1m0s1ty
Err.. nvm
anonymous
  • anonymous
I do not.
4n1m0s1ty
  • 4n1m0s1ty
The information is already given. The slope between t = 3 and t =4 is the acceleration. So whats the acceleration in that time frame equal to (look at the accel graph)
anonymous
  • anonymous
-3
4n1m0s1ty
  • 4n1m0s1ty
ok so the equation for constant acceleration is this: \[a = \frac{ v_{f} - v_{i} }{ t_{f} - t_{i} }\] vf is the final velocity vi is the initial velocity tf is the final time ti is the initial time
4n1m0s1ty
  • 4n1m0s1ty
and a is the acceleration
4n1m0s1ty
  • 4n1m0s1ty
so we know a is -3, vi is 5m/s, ti is 3s, and tf is 3.41s now we just need to solve for vf to find our velocity at 3.41 s
anonymous
  • anonymous
(2m/s-5m/s)/(4s-3s)
4n1m0s1ty
  • 4n1m0s1ty
that would give you the acceleration, but we already found it from the graph.
anonymous
  • anonymous
Right. It's -3
4n1m0s1ty
  • 4n1m0s1ty
so now we need to find the velocity at t = 3.41, \[-3 m/s^{2} = \frac{ v_{f} - 5 m/s}{ 3.41 s - 3 s }\] so solve for vf
anonymous
  • anonymous
4.87
anonymous
  • anonymous
*4.88
4n1m0s1ty
  • 4n1m0s1ty
I'm getting a different answer, I would check your math. Are you using -3 for the acceleration?
anonymous
  • anonymous
Yes
4n1m0s1ty
  • 4n1m0s1ty
Are you doing this: \[(-3m/s^{2})*(3.41 s - 3 s)+5 m/s = v_{f}\] \[(-3)(0.41)+5=v_{f}\] \[-1.23 + 5 = v_{f}\]
anonymous
  • anonymous
I see what I did now.
4n1m0s1ty
  • 4n1m0s1ty
Great
anonymous
  • anonymous
I got 3.77 the second time I did it.
4n1m0s1ty
  • 4n1m0s1ty
yep thats what i got
anonymous
  • anonymous
Thanks for helping out.
4n1m0s1ty
  • 4n1m0s1ty
NP

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