Hello r9k Stupid anon from last week who was here asking for

>>36212706
>cant evaluate the integral of a function composed of straight lines on a fucking grid
I'd tell you to go to /sci/, but they'd also call you retarded.

Brainlets should be culled

>>36212706
go back to escuela pablo

>>36212706
The average value for a straight line is ( f(final t) - f(initial t) ) / 2

Compute this and see they're the same.

>>36212756
/sci/ specifically says not to ask for help with homework


>>36212799
Bitch I'm trying

>>36212806
( f(final t) - f(initial t) ) / 2 + f(initial t), sorry

I'm still not fucking getting it
Goddamn

The only time I get numbers that are the same when I try to do this with the numbers over there is when t = 3 but that's for all I know a coincidence so how am I supposed to look at this?

>>36213157
750 - 150 = 600
600 / 2 = 300
300 + 150 = 450

Are you supposed to do this with integrals or with simple straight line math?

>>36213195
I honestly don't know
I was given this book and this problem is the first one under a paragraph named "scalar kinematics"
It's not a very good book for explaining things, I just have it for the problems

>>36213195
Hmm
Why the + 150?

>>36212756
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YOU'RE JUST AS BAD AS CHADS BORN WITH GREAT FACIAL GENETICS

GO CULL YOURSELF YOU UGLY FUCKING FUCK

RRRRRRRRRRRREEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

>>36213318
It's okay anon
Just like ugly fucks can get plastic surgery to get better us brainlets can study to get better
Don't mind him

>>36213285
750 - 150 gives you the distance between the two points (start and end of the line)
600 / 2 gives you the distance from each point to the average value, therefore
150 + 300 gives you the average value ( and same for 750 -300 )

>>36213368
Oh I think my problem was the 2
I was getting a wrong value cause I was doing 600/4
But that would be the case only if the time interval was counting the whole thing from 0 to 8 instead of just 0 to 4 like it was asked
Right?
If not then I'm not sure why I fucked up

>>36213263
just use the kinematic equations man. if their distance traveled within a time interval is the same, it follows that their average speeds are equivalent. so compute their distances traveled using a kinematic equation

>>36213453
It will always be 2 because you want the midpoint between the start and the end. The halfway point is in the middle, hence / 2. Think graphically.

>>36213492
If this information wasn't given in graph form then how would it work?

>>36213368
So all you need the time for to calculate this was to know which part of the graph they want you to look at?

>>36213368
Isn't the average speed just distance/time?