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Prove that distance travelled by a body ...

Prove that distance travelled by a body in the `n ^(th)` second is ` S _(n) = u + (a)/(2) ( 2n -1)`

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The v-t graph for uniformaly acceerated motion is as shown in the Let `v _(n -1) and v _(n)` be the velocities at t = n -1 and t = n.
Distance travelled by the particle in `n ^(th)` second = area of treapezium `PQST = (1)/(2) (PT + QS) xx TS`
or `S_(n^( th)) = (1)/(2) (v_(n -1) + v _(n))xx [ n - ( n -1)]`
`= (1)/(2) ( v _(n -1) +v_(n)) " "...(i)`
We know that, `v = u + at`
`therefore v _(n -1) = u + a ( n -1) and v _(n) = u + an`
Substituting value of `v _(n -1) and v _(n)` in (i) we get
`S _(n ^( th))= (1)/(2) [ u + a ( n -1) + u + an ]`
`= (1)/(2) ( 2u ) + (1)/(2) (2 an - a)`
`or S _(n ^( th)) = u + (a)/(2) (2n -1)`
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