A particle travels 10m in first 5 sec and 10 m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec.

Options

A. 8.3 m
B. 9.3 m
C. 10.3 m
D. None of above

Answer

A

Explanation

Let `u to ` initial velocity , `S = ut + 1/2 at^2`
a ` to ` accleration
t = 5sec
`10 = u xx 5 + 1/2 xx (a) (5)^2 rArr 10 = 5u + (25a)/(2)` ………..(i)
t = 8sec` to S = 10 + 10 = 20 m`
`20 = u xx 8 + 1/2 xx a xx (8)^2 rArr 20= 8u + 32a`
(i) `rArr 4 = 2u + 5a`
`(ii) rArr 5 = 2u + 8a`
By solving equation
`a = 1/3 m//s^2 , 4 = 2u + 5 xx 1/3`
`rArr u = 7/6 m//s`
Distance travelled in next 2 sec
`=S_10 - S_8`
`= [u xx 10 + 1/2 a xx (10)^2] - 20`
` = [70/6 + 100/6] - 20 = 28.3 - 20 = 8.3m`