Assuming the acceleration in example `2(5.34xx10^(-9)m//s^(2))` above remains constant, how long will Mahendra take to move 1 cm towards Virat?

Explanation

Given :
Distance (s) =1 cm `10^(-2)m`
Acceleration (a) `5.34xx10^(-9)m//s^(2)`
Initial velocity (u) `=0m//s`
To Find : Time taken (t)
Formula: `s=ut+(1)/(2)at^(2)`
Solution :
`s=ut+(1)/(2)at^(2)`
`10^(-2)=(0)t+(1)/(2)(5.34xx10^9-9)t^(2)`
`t=sqrt((2xx10^(-2))/(5.34xx10^(-9)))`
`thereforet=sqrt((1xx10^(-2))/(2.67xx10^(-9)))`
`=sqrt(0.374xx10^(7))`
`=sqrt(374xx10^(4))`
t = 1935 seconds
Mahendra will take 1935 seconds to move 1 cm towards Virat.