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A body sliding on a smooth inclined plane requires 4second to reach the bottom after starting from rest at the top. how much time does it take to cover one

fourth the distance starting from the top


U=0m/s, t=4sec, a=9m/s^2 (acceleration due to gravity )s=ut+1/2at^2=0×4+1/2×9×4×4=72m, 1/4of72=18m Now, s=18m, u=0m/s, a=9m/s^2 s=ut+1/2at^2=>18=0×t+1/2×9×t^2=>18=1/2×9t^2=>t^2=2×2, hence, t=2
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S = u t + 1/2 a t²

a = g Sin Ф,  where Ф = angle of the inclined plane   
u = 0

Hence, s is directly proportional to t².         If distance is 1/4 th  then time taken will be 1/2.

  So  half of 4 secs  is 2 seconds.
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