How long will mahendra take to move 1cm towards virat , if a cosntant acceleration of `5.34xx10^(-9) m//s^2` is produced in mahendra ?

Explanation

Given : Height (s) = 1cm = 0.01 m,
Initial velocity (u) = 0m/s
Acceleration (a) `=5.34xx10^(-9) m//s^2`
To find : Time taken (t)
`therefore s=ut+(1)/(2)at^(2)=0+(1)/(2)at^(2)`
`therefore t=sqrt((2s)/(a))=sqrt((2xx1xx10^(-2)m)/(5.34xx10^(-9)m//s^(2)))`
`=sqrt(0.3745xx10^(7)s^(2))=sqrt(3.745xx10^(6))s`
= 1936s