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 these

Answer

A

Explanation

Let initial ` (t=0 ) ` velocity of particle =u
For first 5 sec ` s_5 =10 m" "rArr " " 10 =5u +(1)/(2) a ( 5)^(2) `
`rArr 2u + 5a =4" "…(i) `
For first 8 sec ` s_B =20 m`
` 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 =uxx 10 +(1)/(2) a (10 )^(2) `
By substituting the value of u and a we get `S_10 = 28.3 m `
So the distance in last ` 2 sec =s_10 -s_B =28 .3 -20 =8.3 m`