OpenStudy (maheshmeghwal9):

OpenStudy (maheshmeghwal9):

OpenStudy (maheshmeghwal9):

OpenStudy (unklerhaukus):

i dont have the lastes version of word, i can not read that document.

OpenStudy (maheshmeghwal9):

@UnkleRhaukus

OpenStudy (unklerhaukus):

integrate?

OpenStudy (maheshmeghwal9):

? i don't know how to integrate:/

OpenStudy (unklerhaukus):

hmm,

OpenStudy (maheshmeghwal9):

ya i don't knw:?

OpenStudy (maheshmeghwal9):

u give me hint or formula i will try:)

OpenStudy (unklerhaukus):

im not sure

OpenStudy (maheshmeghwal9):

np:)

OpenStudy (maheshmeghwal9):

@KingGeorge @satellite73 @saifoo.khan Plz help:)

OpenStudy (maheshmeghwal9):

Its answer is given: - $S=\frac{a}{2 \alpha}.$

OpenStudy (saifoo.khan):

I knew the formula but i forgot. :/ That link might help.

OpenStudy (maheshmeghwal9):

k. thanx:)

OpenStudy (saifoo.khan):

Np! (:

OpenStudy (anonymous):

u can find distance by using ut+1/2at2

OpenStudy (maheshmeghwal9):

I m nt getting Plz u find:) @Ruchi.

OpenStudy (maheshmeghwal9):

Are u getting $S= \frac{a}{2 \alpha}?$ then u right 100% then plz tell me:)

OpenStudy (vincent-lyon.fr):

Hi Mahesh! Distance travelled is distance out + distance return. Here, both distances are equal. So you only need to find distance out, which is equal to displacement from starting point when velocity becomes zero. You will find $$S= 2\Large \frac{a}{4 \alpha}=\frac{a}{2 \alpha}$$? Is that clear?

OpenStudy (maheshmeghwal9):

k! thanx a lot sir:) but wt did u do to find $\frac{S}{2}=\frac{a}{4 \alpha}.$?

OpenStudy (maheshmeghwal9):

only give me hint sir:)

OpenStudy (vincent-lyon.fr):

You get time of U-turn when you write velocity = 0 Then plug it in in s(t)

OpenStudy (maheshmeghwal9):

k. sir thanx a lot lot:)

OpenStudy (vincent-lyon.fr):

yw :)