Explanation:
The ship A is 100 Km west of ship B. The speeds of the ship A and B are 25 Km/hour and 35 Km/hour and the distance between the two ships is 100 Km.
Note: Refer the fig attached
At t hours, the ship A moves at a distance 25 t Km and the ship B move at a distance of 35 t.
The total distance travelled by ship A and ship B is
[tex]25 t+35 t=60 t[/tex]
Note: Refer the 2 image
From the figure, it is observed that the total distance travelled by the two ships is,
[tex]\begin{aligned}L &=\sqrt{100^{2}+(60 t)^{2}} \\&=\sqrt{10000+3600 t^{2}}\end{aligned}[/tex]
Differentiate on both sides with respect to t .
[tex]\begin{aligned}\frac{d L}{d t} &=\frac{d}{d t} \sqrt{10000+3600 t^{2}} \\&=\frac{1}{2 \sqrt{10000+3600 t^{2}}} \frac{d}{d t}\left(10000+3600 t^{2}\right) \\&=\frac{1}{2 \sqrt{10000+3600 t^{2}}}(0+3600(2 t)) \\&=\frac{7200 t}{2 \sqrt{10000+3600 t^{2}}}\end{aligned}[/tex]
At 4 pm (Plug t =4), the speed in which the distance between the two ships is,[tex]$\(\frac{d L}{d t}\right)_{t=4} &\\=\frac{7200(4)}{2 \sqrt{10000+3600(4)^{2}}} \\ &\\=\frac{720}{13} \quad\\\\=55.3846 \mathrm{km} / \mathrm{h}\\=55.4 \mathrm{km} / \mathrm{h} \end{aligned}$[/tex]
