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S = ut + (1/2)at2 ............................(1)
S = ut - (1/2)bt2 ...................................(2)
Differntiating eqn. (2) =
dS/dt = u - (1/2).2bt
u - bt = 0
t = u/b ..........................................(3)
Putting t = u/b in eqn. (2)
we get, S = u2/2b or x = u2/2b or u = underroot(2bx)
Similarly u = underroot2ax/3 is for acceleration.
Hence the condition obtained for covering ''x'' meters in minimum time with max. acc. and max. retard. are:-
u = underroot(2bx) and u = underroot2ax/3 is for acceleration.
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