A community for students.
Here's the question you clicked on:
 0 viewing
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?

This Question is Closed

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1True 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

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1The 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.