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A particle initially at rest, moves with an acceleration 5 m s^(-2) for 5 s. Find the distance travelled in (a) 4 s, (b) 5 s and (c) 5th second. |
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Answer» Solution :Given initial velocity U =0 acceleration a= 5 `m s^(-2)` (a) Distance travelled in t =4 s, `S_(1) =ut +(1)/(2) at^(2) =0xx4 +(1)/(2) xx5 xx(4)^(2)` = 40 m (b) Distance travelled in t = 5 s , `S_(2) = ut +(1)/(2) at^(2) = 0 xx5 +(1)/(2) xx5 xx(5)^(2)` = 62.5 m (c ) Distance travelled in `5^(th)` second , = Distance travelled in 5 s - Distance travelled in 4 s = `S_(2) - S_(1) = (62.5 -40) m = 22.5 m` |
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