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saifoo.khan
 4 years ago
Quick Question, Area under speedtime graph is the distance travelled??
T or F?
saifoo.khan
 4 years ago
Quick Question, Area under speedtime graph is the distance travelled?? T or F?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0True speed = distance/time *time = distance

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0umm is this a physics question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Distance to me is an absolute value. No directional sense. So, as I pointed in the link and as MrYantho suggested, I like displacement.

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.1Okay!! (: can u solve the question in the picture?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Saifoo  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.

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.1No worries. Thanks for trying! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorries me cant do it either :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0um its just at constant speed not instantaneous right?

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.1first it has increasing speed, then constant speed, i want to draw the distancetime graph for this.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0could you wait a bit ill try to help this stuff is in my physics book

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the formula you just use is a regular d=r/t

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.1idk. this is the screenshot form the marking scheme.. i have no idea how they did that.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0formula for aceleration final speed= initial speed + speed change

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.1\[a = \frac{vu}{t}\]Right?

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.1But how we will draw the speedtime? :l

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0on you graph y is the speed x is the time

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.1wait. sorry.. that is already drawn in the question.. i want displacement time. hehe,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0displacement ? umm i dont understand displacement from what?

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.1read the question again, that might help you the part (i) http://assets.openstudy.com/updates/attachments/4f1a4dd4e4b04992dd22a68dsaifoo.khan1327124195861dscn4553.jpg

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you type the rest it?

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.1The first part says to sketch the region whose area represents and stuff.. t And the cut off part says "find the value of T"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first time change of 1 cyclist s=(v) (t) + 1/2 ( a)( t) use formula again to find displacement again of the cyclist

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then you may probably use d=rt to figure out the rest of the stuff i think this is how it is done :\

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's true that the area under speedtime graph is distance you can get this from this formula d=vt dw:1327401093475:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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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