Mathematics: 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 ?

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At noon, ship a is 100km due east of ship b.

Horse0809

31.12.2021

at 4:55 pm

About 64 km

Step-by-step 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= $speed\times times$

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

$\frac{dD}{dt}=200t+2(100-12t)\cdot (-12)$

$\frac{dD}{dt}=0$

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

488t-2400=0

$t=\frac{2400}{488}=4.92 hours$

Again differentiate w.r.t

$\frac{d^2D}{dt^2}=200-24(-12)=200+288=488 0$

Hence, the distance is minimum

The time will be 4:55 pm

Substitute t=4.92

Then AB= $\sqrt{(49.2)^2+(40.96)^2}$

AB=64.01 Km

AB=64 km

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

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