Question

Any physics expert can solve this i will mark it brainliest . A body moving with uniform retardation covers 3km before its speed is reduced to half of its initial​ value .It comes to rest in another distance of -- ans is 1

Answer 1

let initial v be V
so final = v/2 
from eq of motion = v ka square - u ka square = 2as
v/2 ka whole square - v ka square = 2a * 3
after it come to rest final be v = 0
initial become v/2
so from eq of motion = v ka square - u ka square =2as
0 ka square - v /2 ka square = 2as 
0 ka square - v/2 ka whole square =  2as
divide both eq 
u will get s/3 = 1/3 
so s= 1 km

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