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4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Ok this problem is pretty much the same as the last one.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.0So would it start off at 2m/s?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Now what is the velocity at t = 1?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1The acceleration was 2 m/s2 at t = 0, so after 1 second, we give a boost to the initial 2m/s velocity.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So would it be 4 since we are adding 2 to the 2m/s of t=0?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Nope. What's the acceleration at t = 1?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1It goes from 2 m/s2 to ?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1so now for t = 3. Look at the acceleration at t =2 and add is to your velocity.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then would t=3 be 0 since it's a line?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1yes the acceleration is 0.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1so that means the velocity doesn't change.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would the be negative numbers since the acceleration is in the negative now?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.1Are you saying why not v = 2 at t = 3?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Because the acceleration is 3 right not 2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry. Was getting confused with the +2

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1ok so what is the velocity at t = 4

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1the acceleration is back to 0 again right?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1So i think thats it. Any questions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so 0s=2 1s=4 2s=5 3s=0 4s=2 5s=2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry it took so long to type that. My browser kept freezing up on me.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.13s = 5, the acceleration was 0, not the velocity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so if we took the velocity at 3.41s how do we go about doing that?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1look between t = 3 and t = 4 on the velocity graph you just drew.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looks to be about 3.50 to me...does that seem reasonable?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Maybe, 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
 one year ago
Best ResponseYou've already chosen the best response.1Just for my information is this a high school or college physics course?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0He didn't. It's college but this is the first time I've ever had physics.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Ok well I guess its just an estimate then.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It didn't take 3.50 so I think it's wanting more exact.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Ok 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
 one year ago
Best ResponseYou've already chosen the best response.1The 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)

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1ok 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
 one year ago
Best ResponseYou've already chosen the best response.1and a is the acceleration

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1so 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

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1that would give you the acceleration, but we already found it from the graph.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1so 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

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1I'm getting a different answer, I would check your math. Are you using 3 for the acceleration?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Are 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
 one year ago
Best ResponseYou've already chosen the best response.0I see what I did now.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 3.77 the second time I did it.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1yep thats what i got

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for helping out.
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