Question

One ship sailing east with a speed of 7.5m/s passes a certain point at 8A.M. and a second ship sailing north at the same speed passed the same point at 9:30 A.M. at what time are they closest and what is the distance between them then?

Answer 1

Answer:

They are closes at 8:45 AM

The distance between them is 25.64 km

Explanation:

If first ship travels distance A in the East direction and the second ship travels distance B in  the north direction then the distance between them at any instant will be given by C

By pythagoras theorem

$C^2=A^2+B^2$   .............(1)

Let at 8 o'clock time t=0

Thus in time t hour the fist ship will travel distance = 7.5t

or $A=7.5t$

The second ship crosses the point at 9:30 i.e. 1.5 hours after the first ship therefore it will travel distance B in t-1.5 hours

Thus

$B=7.5(t-1.5)$

Thus from (1)

$C^2=(7.5t)^2+[7.5(t-1.5)]^2$

For minimum C

$\frac{dC^2}{dt}=0$

$\implies 2\times 7.5^2t+7.5^2\times 2(t-1.5)=0$

$\implies t+(t-1.5)=0$

$\implies 2t=1.5$

$\implies t=0.75$ hour

or, $t=0.75\times 60$ minutes

or, $t=45$ minutes

Thus the ship will be closest after 45 minutes starting from 8 o clock

i.e. they will be closest at 8:45 AM

The closest distance between them will be

$C^2=7.5^2t^2+7.5^2(t-1.5)^2$

$C^2=7.5^2\times (\frac{18}{5})^2\times [0.75^2+(0.75-1.5)^2]$

$\implies C^2=7.5^2\times (\frac{18}{5})^2\times 2\times 0.75^2$

$\implies C^2=820.125$

$\implies C^2=820.125$

$C=28.64$ km

Hope this helps.