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As the absolute value of slope increases, the velocity (or speed) increase. Should I do it like this?
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.
FInd the integral or the area under the curve and you will have total distance.
Callisto speed and velocity are different don't confuse them. The absolute value is speed and without absolute value you get velocity.
Velocity is a vector speed is a magnitude.
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 = (6-9)/(8-0) = -0.375 For t=8 -11s velocity = (0-6)/(11-8) = -2 Take the absolute value You'll see the the rate of travelling increases.
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.
Please see the axis, it's y=distance , x=time, it's NOT y=velocity , x=time!
in this case velocity is Dx/dt = velocity. In the x direction.
It's the same it's only giving you the distance throughout the total time.
Stop and think. If the line is the rate at which the person is moving. Then the area under the rate is distance.
Perhaps you should read this http://answers.yahoo.com/question/index?qid=20080906075747AABsnoO
Yes view said article answer under the chosen answer, that guy is correct. It is total distance traveled since there is no negative distance.
Displacement is when you have both negative and positive distances traveled. Please hop off my case bro. I'm too pro.
@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?
Displacement as a straight line yes.
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
f(t) = 1/2 t + 9 ; t = [0,8) = 2t + c ; t = (8,10] havent the time to determine "c" tho
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.
@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