An object moving with uniform acceleration covers 40 m in initial 5 seconds and 65 m in next 5 seconds, then its initial velocity is .......

An object moving with uniform acceleration covers 40 m in initial 5 se

1.	An object moving with uniform acceleration covers 40 m in initial 5 seconds and 65 m in next 5 seconds, then its initial velocity is .......
Answer» `4m//s` `2.5m//s` `5.5 m//s` `11m//s` Solution :Displacement in initial 5 SECOND, ` S _(1) = v _(0) t + 1/2 at ^(2)""…(1)` velocity after 5 seconds, `v=v _(0) + at` `THEREFORE` Displacement in next 5 seconds `S_(2) = vt + 1/2 at ^(2)` `S _(2) = (v _(0) + at) t + 1/2 at ^(2)` `therefore S_(2) = v _(0) t + at ^(2) + 1/2 at ^(2) ""...(2)` `therefore` By substracting equation (1) from equation (2), `S_(2) -S_(1) =at ^(2)` `therefore a = (S_(2) - S_(1))/(t ^(2)) = (65- 40)/((5) ^(2)) = (25)/(25) = 1m s^(-2)` From equation (1), In `S_(1) = v_(0)t + 1/2 at ^(2), S_(1) = 40 m, t = 5S and a =1 m//s ^(2),` `40 = 5v _(0) + 1/2 xx 1 xx 25` `therefore 40-12.5 =5v_(0)` ` therefore 27.5 =5v_(0)` `therefore v _(0) =5.5 m //s`