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

- saifoo.khan

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

t

- dumbcow

True
speed = distance/time *time = distance

- anonymous

umm
is this a physics question?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- saifoo.khan

No, this is Math,

##### 1 Attachment

- anonymous

okies :)

- saifoo.khan

Now it's fine.
:D

- anonymous

Distance 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

Okay!! (:
can u solve the question in the picture?

- anonymous

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.

- saifoo.khan

No worries. Thanks for trying! :)

- anonymous

sorries me cant do it either
:(

- saifoo.khan

np

- anonymous

um its just at
constant speed
not instantaneous right?

- saifoo.khan

first it has increasing speed, then constant speed,
i want to draw the distance-time graph for this.

- anonymous

could you wait a bit
ill try to help
this stuff is in my physics book

- anonymous

its the math it uses

- saifoo.khan

This might help,

##### 1 Attachment

- anonymous

the formula you just use is a regular d=r/t

- saifoo.khan

idk. this is the screenshot form the marking scheme.. i have no idea how they did that.

- anonymous

formula for aceleration
final speed= initial speed + speed change

- saifoo.khan

\[a = \frac{v-u}{t}\]Right?

- saifoo.khan

But how we will draw the speed-time? :l

- anonymous

on you graph
y is the speed
x is the time

- saifoo.khan

wait. sorry..
that is already drawn in the question.. i want displacement time. hehe,

- anonymous

displacement ?
umm i dont understand
displacement from what?

- saifoo.khan

read the question again, that might help you
the part (i)
http://assets.openstudy.com/updates/attachments/4f1a4dd4e4b04992dd22a68d-saifoo.khan-1327124195861-dscn4553.jpg

- anonymous

it is cut off

- anonymous

can you type the rest it?

- saifoo.khan

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"

- anonymous

first time change of 1 cyclist
s=(v) (t) + 1/2 ( a)( t)
use formula again to find displacement again of the cyclist

- anonymous

then you may probably use
d=rt
to figure out the rest of the stuff
i think this is how it is done
:\

- anonymous

it's true that the area under speed-time graph is distance
you can get this from this formula d=vt
|dw:1327401093475:dw|

- dumbcow

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.