anonymous
  • anonymous
how do you find average velocity,average speed and do a motion map?
Physics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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theEric
  • theEric
I'm not sure about a motion map. I've never learned about one. But I know equations for average velocity and speed... Do you know the difference between velocity and speed?
anonymous
  • anonymous
Nope
theEric
  • theEric
Okay! Well, when we're talking about speed, we don't care about direction, but we do with velocity. It makes sense that roads have a \(\sf speed\) limit. They signs don't tell you which DIRECTION you have to go!! Haha! When we talk about velocity, we say things like 50 miles per hour \(\sf north\ west\). Or some other direction.

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anonymous
  • anonymous
Ok so like on a vector,can i just attach my homework worksheet...
theEric
  • theEric
Okay! You can take a picture, put it onto your computer, and use the "Attach File" button below the typing box. Velocity is a vector, while speed is a scalar, if that's what you were getting to! :)
anonymous
  • anonymous
Lol yeah that's what i was trying to say,it kind of sucks not knowing this
theEric
  • theEric
It takes a while to understand it :) Take your time, and learn it when you can! So, do you know that distance is a scalar, and displacement is the vector version of distance?
anonymous
  • anonymous
No actually i didnt know that,sorry its taking so long for me to attach my homework my webcam made it look backwards
theEric
  • theEric
Haha! That's okay! Mine can change that in the settings. But, when you have the backwards picture, can you flip it around with the image viewer?
anonymous
  • anonymous
I dont think so..lol now its just not even working let me try and find an online version of the worksheet
theEric
  • theEric
Okay! Can you post the backwards version?
anonymous
  • anonymous
Took it from my phone and got it on here,sorry if its bad quality
theEric
  • theEric
Okay! I can't read any of it, but it seems like the movement is all in one direction or the opposite directoin, huh?
anonymous
  • anonymous
I'm sorry,lol i dont even know anymore but i could try the webcam again
theEric
  • theEric
Well, let me let you know how to find the averages! But first, do you know what the average of some numbers is?
anonymous
  • anonymous
Where you add up all the numbers and divide by however many numbers there are
theEric
  • theEric
Right! So with speed, you add up all of the distances (no direction, so no negative numbers) and divide by the time it took. With velocity, you add up all of the displacements (so the distances can be positive or negative, depending on direction) and divide by the the time it took!
theEric
  • theEric
You don't need to attach the pictures if you learn how to do it in general. Learning how to do it might be easier than getting a good picture, anyway! Haha!
theEric
  • theEric
So what if I said that there is a bee that is flying. First it flies 3m to the right in 4s, and then 4m to the left in 5s. So, try to find the average speed. The total time is the sum of the individual times.
anonymous
  • anonymous
Ok,and what about finding displacement,i tried using the webcam but the whole image was blurry so i broke it up into parts and the parts are not in order so you might have to figure it out,im sorry
theEric
  • theEric
That was a good idea, but they still look a little blurry! And backwards, but I think the Paint program could change that...
theEric
  • theEric
Can you type any of it out?
anonymous
  • anonymous
I dont know because a lot of it is graphing
theEric
  • theEric
|dw:1377828232679:dw|Is that sort of what it looks like?
anonymous
  • anonymous
ok so i found the online worksheet but it gives you the answers but i need to know how to do this and not just get the answers
theEric
  • theEric
That's a good goal :) So, can you give me a link to it, or attach it?
anonymous
  • anonymous
yes i can http://www.highland.k12.in.us/cms/lib04/IN01001438/Centricity/Domain/448/U2_ws_4_solutions.pdf
theEric
  • theEric
Okay! So, shal we start from the top?
anonymous
  • anonymous
okay
theEric
  • theEric
So, I thin I see what the motion map is. Like, each arrow comes from where the object is. The arrow shows its direction, and the length shows its speed! Or, the positions are separated by time intervals. So it would be like if you watched a person walking, and took a picture every second.|dw:1377829187055:dw|
anonymous
  • anonymous
I would think so but i my notes it says arrows represent velocity
theEric
  • theEric
Okay! :) So, since the arrows are the same length, the velocity is staying the same, right?
anonymous
  • anonymous
well apparently according to my notes,the the velocity vector points in the same direction as the displacement
theEric
  • theEric
I think the lines on the motion map are for distance, and there is "1 unit of time" between the circles.
theEric
  • theEric
That is very true, your last statement! :)
anonymous
  • anonymous
Yea in my notes it says for most motion maps the time interval will be one second
theEric
  • theEric
Okay! Sounds good! And maybe the distance marks are every 1 meter. Is that right?
anonymous
  • anonymous
I think so....
theEric
  • theEric
Okay! So, you see where the moving dot starts. in the next picture of it, is it 1 second away. How far did it go?
anonymous
  • anonymous
2 seconds
theEric
  • theEric
2 meters, you probably mean!
anonymous
  • anonymous
lol yes my bad sorry
theEric
  • theEric
In 1 second, it went 2 meters. And then it went 2 more meters in the next second. And then it went 2 more meters in the third second.
theEric
  • theEric
Not a problem! :)
anonymous
  • anonymous
So is it a constant velocity of 2 meters?
theEric
  • theEric
2 meters per second, yep! Now, do you see how there are points on the motion map? Those dots?
anonymous
  • anonymous
Yes the ones that are the starting points for the arrows?
anonymous
  • anonymous
hello??
theEric
  • theEric
Exactly those! I don't know how you assume the initial velocity is zero, but the velocity is 2 meters per second (a.k.a. 2 m/s) to the right. I guess the answer assumes that "right" is the positive direction. Now those points, lets put them on the \(x\) versus \(t\) graph. Each point is at a distance away and a time away from where the dot started. Let's say that the first dot is at zero meters and zero seconds. Then the next dot is at 2 meters and 1 second. Do you see why I am thinking of those numbers? We'll use this soon.|dw:1377830528592:dw|
anonymous
  • anonymous
Yes,I kind of do because one dot represents 2 meters and the dot is on the one second line because of the time interval
theEric
  • theEric
Yeah, like, you can see that the dot moves 2 meters every second. That is its change. So you add 2 meters to the distance and one second to the time to get to the next point. Or you can count on the motion map.|dw:1377831481409:dw|
anonymous
  • anonymous
So the whole thing if you added it would be 10 meters right (looking at the worksheet)
theEric
  • theEric
Yes! :D And in how many seconds?
anonymous
  • anonymous
10 seconds so it would be 1m/s
theEric
  • theEric
Nope! Each new picture of the dot is after one second! So, how many?
anonymous
  • anonymous
5 seconds??
theEric
  • theEric
Yes! So now you know how to read both distance and time changes!! Congrats! Ready to move on to the graphs?
anonymous
  • anonymous
YAY!! hah thank you,you made this so much simpler than what i "learned".Ok yeah I'm ready
theEric
  • theEric
I know what you mean! My pleasure! Okay, so let's start making that \(x\) versus \(t\) plot...
theEric
  • theEric
So, look at the first dot. It is at 0 meters and 0 seconds, so we should put a point on our plot at (0 seconds, 0 meters).
anonymous
  • anonymous
ok so then i should keep plotting,1,1 2,2 3,3 etc
theEric
  • theEric
Actually, you'll get (0,0), (1,2), (2,4).... And what would the next two points be?
anonymous
  • anonymous
4,6 6,8
anonymous
  • anonymous
3,6 4,8
anonymous
  • anonymous
One of the 2 possibilities ??
theEric
  • theEric
(3,6), and (4,8) are right! So, do you see what you were doing to get those? The first part of the point is the time, and the second is the distance. If you add one second, you add two meters.
theEric
  • theEric
Are you familiar with the equation of a line? \(y=mx+b\)?
anonymous
  • anonymous
Yes im not really really familiar but i do know that its slope
theEric
  • theEric
Right! Well, that's how it looks in math when your points are \((x,y)\). Our points are \((t,x)\), so \(x=mt+b\). The slope is rise over run, or \(\dfrac{\Delta x}{\Delta t}\) (where the \(\Delta\) symbol just means "change in," in case you never saw that before).
theEric
  • theEric
If we change 1 second in time, we change 2 meters in distance. \(\Delta t =1\ second\) and \(\Delta x=2\ meters\). So our slope is \(\dfrac{2\ meters}{second}\).
theEric
  • theEric
That is \(m\), slope.
theEric
  • theEric
So, now, you can find the equation for part d!
theEric
  • theEric
I skipped part c, sorry...
anonymous
  • anonymous
Oh ok,a completely off topic question but after i log out will our conversation still be saved because i want to take notes on it
anonymous
  • anonymous
Its fine
theEric
  • theEric
Okay! Yes, this question will always be here. Even after you log out or close the question, it will be on OpenStudy for anybody to see!
anonymous
  • anonymous
Ok thats good,because i have a test on this stuff but back to the learning because i probably only have like 20 mins or so left before i have to go
theEric
  • theEric
Okay! I should get going soon, too! So, do you understand the thing about the line? \(X=\dfrac{2\ meters}{second}\ t+0\). You're homework just simplifies it by cutting off the units. So \(X=2t\). The \(b\) is \(0\) because the intercept is \(0\).
anonymous
  • anonymous
Yeah i get it, the mathematical equation is x=2t because of y=mx+b and instead of y we substituted x as our unit and the slope is 2 and the substitution for x would be time since its on the x axis but we don't necessarily need to add anything because our starting point was 0
theEric
  • theEric
Very, very good! Great! Let's go back to part c, then. Now, the dot is traveling at 2 m/s all the time. Right? So, just make sure that at all \(t\) values you have, you have the \(v\) at 2 m/s.
theEric
  • theEric
That's why the answer sheet did what it did!
anonymous
  • anonymous
Ok, i get it..Thank you so much,you have helped tremendously
theEric
  • theEric
I'm glad! You're welcome! I'm happy to help someone who works to learn!
anonymous
  • anonymous
Ok bye,again,Thank you
theEric
  • theEric
You're welcome. Good luck!

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