OpenStudy (maheshmeghwal9):

Kinematics Problem with differential: - Please help:)

OpenStudy (maheshmeghwal9):

OpenStudy (maheshmeghwal9):

@UnkleRhaukus Please help:)

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):

http://www.desktopclass.com/education/fafsc/projectile-motion-part-1-f-sc-physics-chapter-3-12.html

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 :)