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A body rises vertically up to a height of 125 m in 5 s and then comes back at the point of projection. Find: (i) the total distance travelled, (ii) the displacement, (iii) the average speed and (iv) the average velocity of the body.

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Given, S = 125 m, t = 5 s
(i) Total distance travelled=S+S=2S
= `2 xx 125` m = 250 m
(ii) Displacement = 0 (since final position is same as initial position). Total distance travelled
(iii) Average speed = `("Total distance travelled ")/("Total time of journey")`
`=(2S)/(2t) = (2xx 125 m)/(2xx5s )`
(iv) Average velocity =0 ( since displacement is zero).
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ICSE-MOTION IN ONE DIMENSION -EXERCISE -2 (C) ( Multiple choice type :)
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  2. The correct equation of motion is :

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