Q. A particle starts with initial speed u and retardation a to come to rest in time T. The time taken to cover first half of the total path travelled is? (the answer is in terms of T)
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Your kinematics eq is, for distance traveled "L" in time "T" is;
L = uT - (1/2)aT^2
Then for distance "L/2" in time "t";
L/2 = ut - (1/2)at^2
Since "L" is found from the first eq ,just solve the quadratic eq for "t";
(a/2)t^2 - ut + (L/2) = 0
When you find "t" in terms of "L" (using the quadratic formula) then replace "L" with the first eq. to get "t" in terms of "T".
L = uT - (1/2)aT^2
Then for distance "L/2" in time "t";
L/2 = ut - (1/2)at^2
Since "L" is found from the first eq ,just solve the quadratic eq for "t";
(a/2)t^2 - ut + (L/2) = 0
When you find "t" in terms of "L" (using the quadratic formula) then replace "L" with the first eq. to get "t" in terms of "T".