AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

The figure below represents the ships A and B.
The time required to travel by ship A from A to C is,
t=‌

xA

vA

t=‌

xA

20

The time required to travel by ship B is,
t=‌

xB

vB

t=‌

xB

10

From above two equations,
‌

xA

20

=‌

xB

10

xA=2xB

The shortest distance is,
AB2=xA2+(200−x2B)2

AB=√(2xA)2+(200−x2B)2

AB=√5xB2−400xB+40000‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I)

For the shortest distance, ‌

d

dxB

(AB)=0

0=‌

1

2√5xB2−400xB+40000

(10xB−400)‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II)
xB=40m
Again differentiate equation (II).
‌

d2

dxB2

(AB)>0at‌x=40m
The distance always greater than zero
Substitute the value in equation (I).
AB=√5×402−400×40+40000
AB=80√5km
The time after which the distance between the ships is
shortest is,
t=‌

xB

10

=‌

40

10

=4hr