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 .......

Options

A. `4m//s`
B. `2.5m//s`
C. `5.5 m//s`
D. `11m//s`

Answer

C

Explanation

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`