Question
A BODY STARTS WITH INITIAL VELOCITY 30m/s & A RETARDATION OF 4m/s2 . FIND THE DISTANCE TRAVELLED BY THE BODY IN 8th SECOND.
ANS 1m
Joshi sir comment
by v = u - at
0 = 30 - 4t implies t = 7.5
it means velocity will become 0 in 7.5 sec.
distance covered in 7.5 sec. = 30*7.5 - 1/2*4*(7.5)2 = 225 - 225/2 = 225/2
distance covered in 7 sec. = 30*7 - 1/2*4*(7)2= 210 - 98 = 112
so distance covered in first half of eighth sec. = 112.5 - 112 = 0.5 m
distance covered in last half of eighth sec. = 0*0.5 + 1/2*4(0.5)2= 0.5
so total distance covered in eighth sec. = 0.5 + 0.5 = 1 meter