a particle can have max. acceleration ''a'' and maximum retardation ''b''. devise a strategy so that it covers exactly ''x'' meters in the minimum time

Question

Vikas TU · Answer

S = ut + (1/2)at 2 ............................(1) S = ut - (1/2)bt 2 ...................................(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 = u 2 /2b or x = u 2 /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. Plz Approve! Plz Approve! Plz Approve!

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a particle can have max. acceleration ''a'' and maximum retardation ''b''. devise a strategy so that it covers exactly ''x'' meters in the minimum time

a particle can have max. acceleration ''a'' and maximum retardation ''b''. devise a strategy so that it covers exactly ''x'' meters in the minimum time

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