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IAmCreature
Group Title
Can someone please help me with this? I dont need an answer i just need some explanation on how to do it i guess :)
1.
Describe what the graph shows.
https://www.connexus.com/content/media/932894142005104123AM1766211605.gif
 2 years ago
 2 years ago
IAmCreature Group Title
Can someone please help me with this? I dont need an answer i just need some explanation on how to do it i guess :) 1. Describe what the graph shows. https://www.connexus.com/content/media/932894142005104123AM1766211605.gif
 2 years ago
 2 years ago

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Callisto Group TitleBest ResponseYou've already chosen the best response.2
As the absolute value of slope increases, the velocity (or speed) increase. Should I do it like this?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
find the slope, that will be your velocity. you have two different slopes here. so two different velocities. from 0 to 8 be velocity 1 and 8 to 10 be velocity 2. calculate them.
 2 years ago

UofIMechEng Group TitleBest ResponseYou've already chosen the best response.0
FInd the integral or the area under the curve and you will have total distance.
 2 years ago

UofIMechEng Group TitleBest ResponseYou've already chosen the best response.0
Callisto speed and velocity are different don't confuse them. The absolute value is speed and without absolute value you get velocity.
 2 years ago

UofIMechEng Group TitleBest ResponseYou've already chosen the best response.0
Velocity is a vector speed is a magnitude.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
To my knowledge, area under this graph is NOT the total distance travelled. Or it's better for me to amend my answer . For t=0s  8s velocity = (69)/(80) = 0.375 For t=8 11s velocity = (06)/(118) = 2 Take the absolute value You'll see the the rate of travelling increases.
 2 years ago

UofIMechEng Group TitleBest ResponseYou've already chosen the best response.0
Area under the graph is equal to the Integral of velocity, which is displacement/ position. Which will give you the total are equalling total distance.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Please see the axis, it's y=distance , x=time, it's NOT y=velocity , x=time!
 2 years ago

UofIMechEng Group TitleBest ResponseYou've already chosen the best response.0
in this case velocity is Dx/dt = velocity. In the x direction.
 2 years ago

UofIMechEng Group TitleBest ResponseYou've already chosen the best response.0
It's the same it's only giving you the distance throughout the total time.
 2 years ago

UofIMechEng Group TitleBest ResponseYou've already chosen the best response.0
Stop and think. If the line is the rate at which the person is moving. Then the area under the rate is distance.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Perhaps you should read this http://answers.yahoo.com/question/index?qid=20080906075747AABsnoO
 2 years ago

UofIMechEng Group TitleBest ResponseYou've already chosen the best response.0
Yes view said article answer under the chosen answer, that guy is correct. It is total distance traveled since there is no negative distance.
 2 years ago

UofIMechEng Group TitleBest ResponseYou've already chosen the best response.0
Displacement is when you have both negative and positive distances traveled. Please hop off my case bro. I'm too pro.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
@UofIMechEng , dont get a bee in your bonnet :) Noone was "on your case". Callisto was just trying to clarify their postion to avoid confusion. And yes, area under the velocity curve is displacement  given that we measure it in a straight line right?
 2 years ago

UofIMechEng Group TitleBest ResponseYou've already chosen the best response.0
Displacement as a straight line yes.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
But to answer the question; im not really sure what the graph actually shows :) it looks to be constant velocities between walking and riding home. Without knowing the route taken, the displacement would seem rather moot to me tho
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
f(t) = 1/2 t + 9 ; t = [0,8) = 2t + c ; t = (8,10] havent the time to determine "c" tho
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.0
The graph shows that at time=0 (call this the start time), you are 9 blocks from home. If you look carefully, at time= 3 minutes the graph crosses 8 blocks. That means you are walking 1 block every 3 minutes. At this rate (the slope of the line) it will take you 27 minutes to get home. But at time = 8.5 minutes you get in a car and get home in a total of 11 minutes. Notice that the car is much faster, and the slope of the line is much steeper.
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 Your equations may confuse Creature. If you magnify the graph you see that the slope is 1/3 on the first line. The 2nd line goes from 8.5 to 11 mins
 2 years ago
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