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.

- anonymous

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- 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

Right

- 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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## More answers

- 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

So what would be his position at time 0?

- anonymous

2m?

- 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

Is that clear or is it still confusing?

- anonymous

Oh

- anonymous

A little more clear but physics kinda confuses me in general.

- 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

I get why it's -1 for the first one but I don't know how you would figure out the rest though.

- 4n1m0s1ty

Ok so next part.

- 4n1m0s1ty

We need to examine the velocity graph

- 4n1m0s1ty

position tells you where you are, and velocity tells you where you are going

- 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

The graph goes positive so does that mean he's going away from it?

- 4n1m0s1ty

It depends. Just looking at whether its positive or negative isn't enough.

- 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

|dw:1444358671721:dw|

- anonymous

so if I was plotting this the next would be -2?

- 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

Or rather -3?

- 4n1m0s1ty

yep

- 4n1m0s1ty

and now you just repeat what we just did for each time.

- anonymous

so the next is -5?

- 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

he would be at - 5 m only if he kept the same velocity from before, but he changes velocity at t = 1/

- anonymous

So t=3 it would be -4 since he's moving again?

- 4n1m0s1ty

Well it would change, but it wouldn't be -4 m

- 4n1m0s1ty

Look at t =3, whats the velocity?

- anonymous

2

- 4n1m0s1ty

I mean, err, t = 2?

- anonymous

It's 0

- 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

Does it make sense?

- anonymous

Yeah..sort of

- 4n1m0s1ty

Is there something confusing that you don't understand? Don't be afraid to ask :-).

- anonymous

So the next is 1?

- 4n1m0s1ty

Yep.

- anonymous

And the last one is 3

- 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

yes

- 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

Sure, if it doesn't take too long. But was there anything confusing about how to solve problem?

- 4n1m0s1ty

* this problem

- anonymous

I just don't really know how to approach these sort of problems.

- 4n1m0s1ty

Well do you understand what position is?

- 4n1m0s1ty

and velocity?

- anonymous

More velocity than position.

- anonymous

We just started this stuff so it's all new to me.

- 4n1m0s1ty

Ok tell me in you words what you think the 2 are? I need to gauge your understanding.

- anonymous

Velocity is the change of the position
Positions is where the object is at?

- 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

It's relative then

- 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

When you are ready just start an new thread.

- anonymous

Will do.

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