A body moving with a uniform acceleration crosses a distance of 15 m in the `3^(rd)` second and `23 m` in the `5^(th)` second. The displacement in `10 s` will be

Options

A. 100 m
B. 80 m
C. 60 m
D. 20 m

Explanation

Let u is the initial velocity and a is the acceleration then
`S_(n) = u + (1//2) a (2n -1) `
` therefore S_(3) = u + (1//2) a (3xx 2 -1)`
` rArr 4 = u + (5)/(2) a ` …(i)
similarly for `5^(th)` second
` S_(5) = u + (1//2) a (2xx 5 - 1)`
` rArr 12 = u + (9//2) a ` ...(ii)
From (i) & (ii) u = - 6 m/s
and a = ` 4 m//s^(2)`
so, distance travelled in 5 sec
From s = ut + 1/2 ` "at"^(2)`
` s = - 6 xx 5 + (1//2) xx 4 xx 5^(2) = 20 m `
Similarly distance travelled in 8 sec
`= –6 xx 8 + (1//2) 4 xx 8^(2) = 80 m`
So distance travelled in next 3 sec
= 80 – 20 = 60 m
Hence correct answer is (C)