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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.
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.)
So if we know that the mailbox is 1 meter in front of john, that means he is 1 m behind the mailbox.
So what would be his position at time 0?
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.
Is that clear or is it still confusing?
A little more clear but physics kinda confuses me in general.
Well thats ok, its hard at first but once you take the time to actually think through it. It makes a lot of sense.
I get why it's -1 for the first one but I don't know how you would figure out the rest though.
Ok so next part.
We need to examine the velocity graph
position tells you where you are, and velocity tells you where you are going
So if we look at the velocity of john at t = 0. We find that he is moving at rate of - 2 m/s.
The graph goes positive so does that mean he's going away from it?
It depends. Just looking at whether its positive or negative isn't enough.
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.
so if I was plotting this the next would be -2?
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.
Or rather -3?
and now you just repeat what we just did for each time.
so the next is -5?
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.
he would be at - 5 m only if he kept the same velocity from before, but he changes velocity at t = 1/
So t=3 it would be -4 since he's moving again?
Well it would change, but it wouldn't be -4 m
Look at t =3, whats the velocity?
I mean, err, t = 2?
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.
Does it make sense?
Is there something confusing that you don't understand? Don't be afraid to ask :-).
So the next is 1?
And the last one is 3
The velocity doesn't change for the rest of the problem, so you just need to add 2 to the position after each time.
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.
Sure, if it doesn't take too long. But was there anything confusing about how to solve problem?
* this problem
I just don't really know how to approach these sort of problems.
Well do you understand what position is?
More velocity than position.
We just started this stuff so it's all new to me.
Ok tell me in you words what you think the 2 are? I need to gauge your understanding.
Velocity is the change of the position Positions is where the object is at?
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).
It's relative then
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).
When you are ready just start an new thread.