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. 9.3m
C. 10.3m
D. 5.6m

Answer

A

Explanation

Let initial (t=0) velocity of particle= u
From first 5 sec motion the distance is, `s_(5) = 10` metre
`s= u t+ (1)/(2) a t^(2) rArr 10 = 5u + (1)/(2)a (5)^(2)" " 2u + 5a =4` …(i)
For first 8 sec of motion the distance is, `s_(8)= 20` metre `rArr 20=8u + (1)/(2) a(8)^(2) rArr 2u + 8a = 5` ...(ii)
By solving `u= (7)/(6) m//s and a= (1)/(3) m//s^(2)`
Now distance travelled by particle in total 10 sec. `s_(10) = u xx 10 + (1)/(2) a (10)^(2)` By substituting the value of u and a we will get `s_(10) = 28.3m`
So, the distance in last 2 sec `= s_(10) s_(8)= 28.3 - 20 = 8.3m`