## IAmCreature 3 years ago 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/93289-4142005-104123-AM-1766211605.gif

1. Callisto

As the absolute value of slope increases, the velocity (or speed) increase. Should I do it like this?

2. experimentX

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.

3. UofIMechEng

FInd the integral or the area under the curve and you will have total distance.

4. UofIMechEng

Callisto speed and velocity are different don't confuse them. The absolute value is speed and without absolute value you get velocity.

5. UofIMechEng

Velocity is a vector speed is a magnitude.

6. Callisto

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.

7. UofIMechEng

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.

8. Callisto

Please see the axis, it's y=distance , x=time, it's NOT y=velocity , x=time!

9. UofIMechEng

in this case velocity is Dx/dt = velocity. In the x direction.

10. UofIMechEng

It's the same it's only giving you the distance throughout the total time.

11. UofIMechEng

Stop and think. If the line is the rate at which the person is moving. Then the area under the rate is distance.

12. Callisto

13. UofIMechEng

Yes view said article answer under the chosen answer, that guy is correct. It is total distance traveled since there is no negative distance.

14. UofIMechEng

Displacement is when you have both negative and positive distances traveled. Please hop off my case bro. I'm too pro.

15. amistre64

@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?

16. UofIMechEng

Displacement as a straight line yes.

17. amistre64

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

18. amistre64

f(t) = 1/2 t + 9 ; t = [0,8) = 2t + c ; t = (8,10] havent the time to determine "c" tho

19. phi

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.

20. phi

@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

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