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A particle starting from initial velocity v_0 moves on straight line with constant acceleration equation of distance travelled during n^"th" second will be …... .

Answer»

SOLUTION :For given particle, DISTANCE TRAVELLED in `n^(th)` seconde is, d = (distance travelled in n seconds)
- (distance travelled in (n-1) second)
`= (v_(0) n + 1//2 an ^(2)) -(v_(0) (n -1) +1/2 a (n-1) ^(2))`
(From equation `d = v _(0) t + 1/2 at ^(2))`
`=( v _(0) n + 1/2 an ^(2)) - (v _(0) n - v_(0) + a//2 (n ^(2) -2n +1))`
`=(v_(0) n + 1/2an ^(2) -v _(0) n + v_(0) -1/2 an ^(2) + an - (a)/(2))`
`= v _(0) + a _(n) - (a)/(2)`
`therefore d = v _(0) + a/2 (2n -1)`


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