anonymous
  • anonymous
Any help? It's velocity and position sort of stuff. http://oi60.tinypic.com/21msxlz.jpg The question is in the link since it's a picture.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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4n1m0s1ty
  • 4n1m0s1ty
ok so lets break this problem down. The first thing you need to find out is where john starts. According to the problem, the mailbox is 1 m in front of him.
anonymous
  • anonymous
Right
4n1m0s1ty
  • 4n1m0s1ty
the problem states that the mailbox is the zero location, meaning that it is designated as position 0 m, and that any position that john takes is relative to that position (e.g. at 2 m he is 2 meters in front of the mailbox, - 4 m is 4 meters behind the mailbox.)

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4n1m0s1ty
  • 4n1m0s1ty
So if we know that the mailbox is 1 meter in front of john, that means he is 1 m behind the mailbox.
4n1m0s1ty
  • 4n1m0s1ty
So what would be his position at time 0?
anonymous
  • anonymous
2m?
4n1m0s1ty
  • 4n1m0s1ty
Not exactly. It would be -1 m. He is 1 m behind the mailbox, so if we use it as a reference point, it would be measured as -1 m.
4n1m0s1ty
  • 4n1m0s1ty
Is that clear or is it still confusing?
anonymous
  • anonymous
Oh
anonymous
  • anonymous
A little more clear but physics kinda confuses me in general.
4n1m0s1ty
  • 4n1m0s1ty
Well thats ok, its hard at first but once you take the time to actually think through it. It makes a lot of sense.
anonymous
  • anonymous
I get why it's -1 for the first one but I don't know how you would figure out the rest though.
4n1m0s1ty
  • 4n1m0s1ty
Ok so next part.
4n1m0s1ty
  • 4n1m0s1ty
We need to examine the velocity graph
4n1m0s1ty
  • 4n1m0s1ty
position tells you where you are, and velocity tells you where you are going
4n1m0s1ty
  • 4n1m0s1ty
So if we look at the velocity of john at t = 0. We find that he is moving at rate of - 2 m/s.
anonymous
  • anonymous
The graph goes positive so does that mean he's going away from it?
4n1m0s1ty
  • 4n1m0s1ty
It depends. Just looking at whether its positive or negative isn't enough.
4n1m0s1ty
  • 4n1m0s1ty
His position at t = 0 is -1 m. His velocity is -2 m/s at t =0. The negative velocity and negative position tells you that he will be moving further from the mailbox.
4n1m0s1ty
  • 4n1m0s1ty
|dw:1444358671721:dw|
anonymous
  • anonymous
so if I was plotting this the next would be -2?
4n1m0s1ty
  • 4n1m0s1ty
Almost. You have to consider his initial position. If 1 second passed he would move away 2 m from the mailbox, but he started at -1 m. So that would make him be at -3 m at t =1.
anonymous
  • anonymous
Or rather -3?
4n1m0s1ty
  • 4n1m0s1ty
yep
4n1m0s1ty
  • 4n1m0s1ty
and now you just repeat what we just did for each time.
anonymous
  • anonymous
so the next is -5?
4n1m0s1ty
  • 4n1m0s1ty
At t = 1, he is at -3 m and his velocity is 0 m, so he doesn't move. So at t = 2, he is at the same position.
4n1m0s1ty
  • 4n1m0s1ty
he would be at - 5 m only if he kept the same velocity from before, but he changes velocity at t = 1/
anonymous
  • anonymous
So t=3 it would be -4 since he's moving again?
4n1m0s1ty
  • 4n1m0s1ty
Well it would change, but it wouldn't be -4 m
4n1m0s1ty
  • 4n1m0s1ty
Look at t =3, whats the velocity?
anonymous
  • anonymous
2
4n1m0s1ty
  • 4n1m0s1ty
I mean, err, t = 2?
anonymous
  • anonymous
It's 0
4n1m0s1ty
  • 4n1m0s1ty
Yeah its 2 m/s right? so if he was at -3m, we add positive 2 m after 1 second. So it would be -1 m.
4n1m0s1ty
  • 4n1m0s1ty
Does it make sense?
anonymous
  • anonymous
Yeah..sort of
4n1m0s1ty
  • 4n1m0s1ty
Is there something confusing that you don't understand? Don't be afraid to ask :-).
anonymous
  • anonymous
So the next is 1?
4n1m0s1ty
  • 4n1m0s1ty
Yep.
anonymous
  • anonymous
And the last one is 3
4n1m0s1ty
  • 4n1m0s1ty
The velocity doesn't change for the rest of the problem, so you just need to add 2 to the position after each time.
4n1m0s1ty
  • 4n1m0s1ty
yes
anonymous
  • anonymous
Okay. I got that one. Would you be able to help with some more? It's still kinda rocky for me but I'm getting it better.
4n1m0s1ty
  • 4n1m0s1ty
Sure, if it doesn't take too long. But was there anything confusing about how to solve problem?
4n1m0s1ty
  • 4n1m0s1ty
* this problem
anonymous
  • anonymous
I just don't really know how to approach these sort of problems.
4n1m0s1ty
  • 4n1m0s1ty
Well do you understand what position is?
4n1m0s1ty
  • 4n1m0s1ty
and velocity?
anonymous
  • anonymous
More velocity than position.
anonymous
  • anonymous
We just started this stuff so it's all new to me.
4n1m0s1ty
  • 4n1m0s1ty
Ok tell me in you words what you think the 2 are? I need to gauge your understanding.
anonymous
  • anonymous
Velocity is the change of the position Positions is where the object is at?
4n1m0s1ty
  • 4n1m0s1ty
Is position relative or absolute? Can I say tell you that I am at 1 meter without any other information and still have that make sense. Also, velocity is the change in position "over time". It is an important distinction because you can have change in position over another variable (outside your course probably). Velocity tells you where you are going (both the speed and direction).
anonymous
  • anonymous
It's relative then
4n1m0s1ty
  • 4n1m0s1ty
Yes, so when w approach problems like these (called kinematics problems). We always establish a frame of reference (or like in this problem have it defined for us e.g. the mailbox).
4n1m0s1ty
  • 4n1m0s1ty
When you are ready just start an new thread.
anonymous
  • anonymous
Will do.

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