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Mathematics 13 Online
OpenStudy (anonymous):

Help @thomaster

OpenStudy (anonymous):

A car travelling at 72km/h decelerates uniformly at 2m/s^2. Calculate (a)the distance it goes before it stops (b)the time it takes to stop (c)the distance it travels during the first and third seconds

OpenStudy (anonymous):

@Mashy

OpenStudy (anonymous):

This is a trivial problem .. what are the initial data given?

OpenStudy (anonymous):

Initial data?

OpenStudy (anonymous):

to solve these problems.. u need the three equations of motion.. right? u know them?

OpenStudy (anonymous):

yes yes

OpenStudy (anonymous):

\[a(t)=-2\] There's one!

OpenStudy (anonymous):

yea.. so like he said.. a = -2 m/s^2.. what else do u know?

OpenStudy (anonymous):

Oh lol acceleration is -2km/h initial velocity = 72km/h

OpenStudy (anonymous):

-2m/s^2 sorry

OpenStudy (anonymous):

yea ok.. so u need to convert it into m/s.. do u know the final velocity? (the car comes to stop)

OpenStudy (anonymous):

and converting that 72km/h into m/s gives 20

OpenStudy (anonymous):

yea final velocity is 0

OpenStudy (anonymous):

ok.. so u have u , v, and a can u calculate t and S ?

OpenStudy (anonymous):

)the distance it goes BEFORE it stops means does it have any significance

OpenStudy (anonymous):

BEFORE = distance travelled till it stops.. its just a way of speaking in english.. not any significance.. :)

OpenStudy (anonymous):

Oh , then it's super-duper trivial

OpenStudy (anonymous):

now that i think about it.. this is really genuine doubt.. which students can have.. and we teachers tend to over look thank you :).. i ll keep this in mind while teaching :) :). for next time

OpenStudy (anonymous):

yes , teachers tend to overlook U a teacher??????

OpenStudy (anonymous):

yup :D :D..

OpenStudy (anonymous):

(c)the distance it travels during the first and third seconds what should we do for this distance travelled in 3 secons - distanc etravelled in 1 second

OpenStudy (anonymous):

?

OpenStudy (anonymous):

during first second distance travelled in first second = distance travelled in two seconds - distance travelled in 1 second in general.. distance travelled in nth second = distance travelled in total n seconds - distance travelled in (n-1) seconds

OpenStudy (anonymous):

What is the difference , why can't we just take 1 second in our equation

OpenStudy (anonymous):

1st second 2nd second or 3rd second.. all are 1 second time intervals.. so u need to know which 1 second interval we are talking about right?

OpenStudy (anonymous):

but yea.. distance in 1 scond = distance in first second.. i agree but distance in 2 second is not equal to distance in 2nd second

OpenStudy (anonymous):

Why

OpenStudy (anonymous):

sry for trouble though

OpenStudy (anonymous):

u teach where btw

OpenStudy (anonymous):

u can think distance travelled in 2 seconds is.. total distance in 2 seconds interval but distance travelled in 2nd second is distance travelled in the 2nd 1 second ka time interval.. lets take simple example consider an object with zero speed.. and accelerates at 2m/s then in 1second it goes s = 1/2 at^2 = 1m in 2 second it goes s= 4m in 3 second it goes s = 9m in 4 second it goes s = 16 m.. and so on but.. now if i ask.. distance travlled in 2nd second.. it is 4 - 1 = 3m.. (this is the distance coverd by the object in 2nd 1 second interval) also distance in 3rd second = 9 - 4 = 5m.. (again.. this is distance travelled in the 3rd one second interval).. get it?

OpenStudy (anonymous):

Oh i got the concept But if the question was , the particle is at what position at 3 rd second then the answer is 9m right

OpenStudy (anonymous):

u can't say.. AT 3rd second cause 3rd second is an interval.. and in that interval.. the particle is travelling from 4m to 9 m.. so if u ask.. the position of the particle at the BEGINNING of 3rd second.. it is at 4m and at the END of 3rd second.. it is at 9 m

OpenStudy (anonymous):

This is confusing but i get it btw where do u teach

OpenStudy (anonymous):

ur a man

OpenStudy (anonymous):

?

OpenStudy (anonymous):

yea m a man :P.. i teach in mangalore..

OpenStudy (anonymous):

grt , if u r searching for questions in physics http://manishkumarphysics.in/kota/?page_id=271

OpenStudy (anonymous):

(b)the time it takes to stop? Do we use v=u+at

OpenStudy (anonymous):

yes.. use that

OpenStudy (anonymous):

thanks for the link.. is that ur site?

OpenStudy (anonymous):

if that was my site i wouldn't have asked this question lol

OpenStudy (anonymous):

that is for IIT preparation

OpenStudy (anonymous):

haha lol.. ok xD

OpenStudy (anonymous):

CAn we do the last part , i feel to discuss it

OpenStudy (anonymous):

@Mashy

OpenStudy (anonymous):

yes.. calculate..

OpenStudy (anonymous):

i am not getting just the last part

OpenStudy (anonymous):

fist second . u put t = 1 right? for third second find distance in 3 second, distance in 2 second.. then subtract

OpenStudy (anonymous):

yes i made a math error

OpenStudy (anonymous):

:P

OpenStudy (anonymous):

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