The two ships are sailing in opposite sides, Hence, the resultant speed is given by 15+5 = 20 km/hr.
Therefore, from the below triangle, we have
[tex]\frac{dx}{dt} = 20 \text{ km/hr}[/tex]
Let the distance between the ships is y. On applying Pythagorous theorem, we have
[tex]y^2=x^2+60^2\\ \text{On differentiating, we get}\\ 2y\frac{dy}{dt} =2x\frac{dx}{dt}\\\frac{dy}{dt}= \frac{x}{y} \frac{dx}{dt}[/tex]
On substituting the value of y as [tex]y=\sqrt{60^2+x^2}[/tex]
[tex]\frac{dy}{dt} =\frac{x}{\sqrt{60^2+x^2}} \frac{dx}{dt}[/tex]
Since, the at noon the ship is 60 km to each other. Hence, for 4 PM, i.e. t=4, we have
[tex]x=4 \times \frac{dx}{dt} \\ x= 4 \times 20 \\x=80[/tex]
On substituting the value in above, we get
[tex]\frac{dy}{dt} =\frac{80}{\sqrt{60^2+80^2}} (20[/tex]
[tex]\frac{dy}{dt} = 16.0 \text{ km/hr}[/tex]
Therefore, the distance between the ships changing at a rate of 16 km/hr at 4:00 pm
