Posted by **anonymous** on Sunday, June 16, 2013 at 9:14am.

a body moves for a total of 9 seconds starting from rest with uniform acceleration and then with uniform retardation which is twice of value of acceleration to stop .what is the duration of uniform acceleration

- Physics 11th -
**Anonymous**, Sunday, August 24, 2014 at 2:07am
v=u+at

- Physics 11th -
**Pritam singh**, Friday, August 29, 2014 at 8:16am
Distance covered in 4 secods after tha 5th seconds= S9- s5 .tha distance covered in t second is given by s = ut 1/2 at2and distance covered in nth second is given by Dn = u a/2 (2n-1)

- Physics 11th -
**Pritam singh**, Friday, August 29, 2014 at 8:18am
Distance covered in 4 secods after tha 5th seconds= S9- s5 .tha distance covered in t second is given by s = ut 1/2 at2and distance covered in nth second is given by Dn = u a/2 (2n-1) The modern sr. sec. Bagpur palwal phone no. 8059892363

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**Genius**, Wednesday, June 10, 2015 at 3:41am
Use v=u+at and form two equations one for acc. And one for retard. Then on solving u will get d ans. I hope dat helps.. 😀 ☺

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