1.

Assuming that acceleration remains constant `(5.34xx10^(-9)m//s^(2))`, How long will Mahendra take to move 1 cm towards Virat if he starts from rest?

Answer» Given Acceleration `(a)=6.34xx10^(-9)m//s^(2)`
`=5xx10^(-9)m//s^(2)`
Displacement `(s)=1cm=1/100m`
Initial velocity `(u)=0ml//s`
To find: time `(t)=?`
Formula`s=ut+1/2at^(2)`
Solution: `1/100=0xxt+1/2xx5xx10^(-9)xxt^(2)`
`1/100=2.5xx10^(-9)xxt^(2)`
`1/(100xx2.5xx10^(-9))=t^(2)`
`(1xx10^(9))/250=t^(2)`
`(1000xx10^(6))/250=t^(2)`
`4xx10^(6)=t^(2)`
`t=sqrt(4xx10^(6)s)`
`t=2xx10^(3)s`
Mahendra will take `2xx10^(3)s` to move towards Virat.


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