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

S = ut + (1/2)at² ............................(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 = 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.

Plz Approve!

Plz Approve!

Plz Approve!