Quick Question, Area under speed-time graph is the distance travelled?? T or F?

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Quick Question, Area under speed-time graph is the distance travelled?? T or F?

Mathematics
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t
True speed = distance/time *time = distance
umm is this a physics question?

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Other answers:

No, this is Math,
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okies :)
Now it's fine. :D
Distance to me is an absolute value. No directional sense. So, as I pointed in the link and as MrYantho suggested, I like displacement.
Okay!! (: can u solve the question in the picture?
Saifoo - sorry man....I did this stuff 22 years ago and it is hard to remember. I can google it up and get it. But, you or someone current is going to be a lot better at it.
No worries. Thanks for trying! :)
sorries me cant do it either :(
np
um its just at constant speed not instantaneous right?
first it has increasing speed, then constant speed, i want to draw the distance-time graph for this.
could you wait a bit ill try to help this stuff is in my physics book
its the math it uses
This might help,
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the formula you just use is a regular d=r/t
idk. this is the screenshot form the marking scheme.. i have no idea how they did that.
formula for aceleration final speed= initial speed + speed change
\[a = \frac{v-u}{t}\]Right?
But how we will draw the speed-time? :l
on you graph y is the speed x is the time
wait. sorry.. that is already drawn in the question.. i want displacement time. hehe,
displacement ? umm i dont understand displacement from what?
read the question again, that might help you the part (i) http://assets.openstudy.com/updates/attachments/4f1a4dd4e4b04992dd22a68d-saifoo.khan-1327124195861-dscn4553.jpg
it is cut off
can you type the rest it?
The first part says to sketch the region whose area represents and stuff.. t And the cut off part says "find the value of T"
first time change of 1 cyclist s=(v) (t) + 1/2 ( a)( t) use formula again to find displacement again of the cyclist
then you may probably use d=rt to figure out the rest of the stuff i think this is how it is done :\
it's true that the area under speed-time graph is distance you can get this from this formula d=vt |dw:1327401093475:dw|
The line for cyclist P is v = 2t, since the constant acceleration is given as 2 the displacement is 16 and the area under the curve is the displacement Area = vt/2 = 16 vt = 32 substitute v=2t 2t^2 = 32 t^2 = 16 t =4 Therefore, cyclist Q starts 4 sec after when cyclist P is going 8 m/s

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