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anonymous
 one year ago
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
 one year ago
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.

This Question is Closed

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

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

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1So if we know that the mailbox is 1 meter in front of john, that means he is 1 m behind the mailbox.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1So what would be his position at time 0?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Not 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
 one year ago
Best ResponseYou've already chosen the best response.1Is that clear or is it still confusing?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A little more clear but physics kinda confuses me in general.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Well thats ok, its hard at first but once you take the time to actually think through it. It makes a lot of sense.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get why it's 1 for the first one but I don't know how you would figure out the rest though.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1We need to examine the velocity graph

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1position tells you where you are, and velocity tells you where you are going

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1So if we look at the velocity of john at t = 0. We find that he is moving at rate of  2 m/s.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The graph goes positive so does that mean he's going away from it?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1It depends. Just looking at whether its positive or negative isn't enough.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1His 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
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444358671721:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if I was plotting this the next would be 2?

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

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1and now you just repeat what we just did for each time.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1At 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
 one year ago
Best ResponseYou've already chosen the best response.1he would be at  5 m only if he kept the same velocity from before, but he changes velocity at t = 1/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So t=3 it would be 4 since he's moving again?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Well it would change, but it wouldn't be 4 m

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Look at t =3, whats the velocity?

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.1Is there something confusing that you don't understand? Don't be afraid to ask :).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And the last one is 3

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1The velocity doesn't change for the rest of the problem, so you just need to add 2 to the position after each time.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay. 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
 one year ago
Best ResponseYou've already chosen the best response.1Sure, if it doesn't take too long. But was there anything confusing about how to solve problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just don't really know how to approach these sort of problems.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Well do you understand what position is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0More velocity than position.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We just started this stuff so it's all new to me.

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Ok tell me in you words what you think the 2 are? I need to gauge your understanding.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Velocity is the change of the position Positions is where the object is at?

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

4n1m0s1ty
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.1When you are ready just start an new thread.
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