A particle starts its motion from rest under the action of a constant force. If the distance covered in first `10s` is `s_(1)` and that covered in the first `20 s` is `s_(2)`, then

Options

A. `s_(2)=2s_(1)`
B. `s_(2)=3s_(1)`
C. `s_(2)=4s_(1)`
D. `s_(2)=s_(1)`

Answer

C

Explanation

A particle ………..
If the particle is moving in a straight line under the action of a constant force then distance covered
`s=ut+(1)/(2)at^(2)`
Since the body starts from rest `u = 0`
`:. s = (1)/(2)at^(2)`
Now, `s_(1)= (1)/(2)a(10)^(2)` ............(i)
and, `s_(2)= (1)/(2)a(20)^(2)` .............. (ii)
Deividing Eq (i) and Eq (ii) we get
`(s_(1))/(s_(2)) = ((10)^(2))/((20)^(2))`
`implies s_(2) = 4s_(1)`