there are two strings attached
ABCD and EDFG
In a time interval t
Let the elevator goes up by a distance x
and Let pully D goes up by a distance y
then,
length AB has been reduced by x
length BC has been reduced by x
length FG has been reduced by x
length CDE has been reduced by x
length CDF has been reduced by x
length CD has been reduced by y ............{As D goes up by y}
∴ length DE has been reduced by x-y
and, length DF has been reduced by x-y
Thus
total decrease in the length ABCD = decrease in AB + decrease in BC + decrease in CD
= x + x + y
= 2x + y
total decrease in the length EDFG = decrease in ED + decrease in DF + decrease in FG
= (x-y) + (x-y) + x
= 3x - 2y
Now, decrease in length ABCD has been drawn in by motor at A
and, decrease in length EDFG has been drawn in by motor at G
speed of motor at A = (2x + y) / t
speed of motor at G = (3x - 2y) / t
according to question,
(2x + y) / t = 5 m/s ............................(1)
(3x - 2y) / t = 5 m/s ...........................(2)
Thus (2x + y) / t = (3x - 2y) / t
or, 2x + y = 3x - 2y
or, 3y = x
or, y = x/3
putting in (1)
(2x + x/3) / t = 5 m/s
or, (x/t)*(7/3) = 5 m/s
or, velocity of elevator = x/t = (15/7) m/s