1.

In the above question what is the distance covered bythe particle in 5th second of its motion ?

Answer»

4 m
8 m
5 m
11 m

Solution :`x=(t-2)^(2)=t^(2)+4-4`
`(dx)/(dt)=v=2t-4`
`:.` Velocity u at t=0is u=-4 `ms^(-1)`
ALSO `(dv)/(dt)=a=2 ms^(-2)`
Thus the body has uniform acceleration.
Here `S_(n)=u+(a)/(2)(2n-1)`
`=-4+2//2(2xx5-1)`
`= -4+9=5 m`


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