Question

At noon, ship A is 100km due east of ship B. Ship A is sailing west at 12km per hour, and ship B is sailing south at 10km per hour. At what time will the ships be nearest to each other, and what will this distance be ?

Answer 1

Answer: After 4.55 hours the ships will be nearest to each other, and this distance would be 54.54 km.

Explanation:

Since we have given that

Distance between ship A and ship B = 100 km

Speed at which Ship A is sailing west = 12 km/hr

Speed at which Ship B is sailing south = 10 km/hr

Since they are sailing opposite to each other, so their relative speed would be

`12+10=22\ km/hr`

So, Time taken is given by

`time=(Distance)/(Speed)=(100)/(22)=4.55\ hrs`

Distance would be

`(x)/(12)=(100-x)/(10)\\\\10x=12(100-x)\\\\10x=1200-12x\\\\10x+12x=1200\\\\ 22x=1200\\\\x=(1200)/(22)\\\\x=54.54\ km`

Hence, After 4.55 hours the ships will be nearest to each other, and this distance would be 54.54 km.

Answer 2

at 4:55 pm

About 64 km

Explanation:

Let t be time in hours taken by ship A to point B and ship B takes time t hours to point A

Speed of ship A=12 km/h

Speed of ship B=10 km/h

Distance=
`speed* times`

Distance travel by ship A=12 t

Distance travel by ship B =10 t

OA=10 t

OC=100 km

OB=OC-BC=100-12 t

Using pythagorous theorem

`(Hypotenuse)^2=(perpendicular)^2+(base)^2`

`AB^2=(10t)^2+(100-12 t)^2`

`D=(10t)^2+(100-12 t)^2`

The distance between two ships is minimum

Differentiate w.r.t time

`(dD)/(dt)=200t+2(100-12t)\cdot (-12)`

`(dD)/(dt)=0`

`200 t-24(100-12t)=0`

`488t-2400=0`

`t=(2400)/(488)=4.92 hours`

Again differentiate w.r.t

`(d^2D)/(dt^2)=200-24(-12)=200+288=488 > 0`

Hence, the distance is minimum

The time will be 4:55 pm

Substitute t=4.92

Then AB=
`√((49.2)^2+(40.96)^2)`

AB=64.01 Km

AB=64 km

Hence, the distance between two ships =64 km (about)