Question

A submarine can travel at 25 knots with the current and at

16 knots against it. Find the speed of the current and the speed
of the submarine in still water.

Answer 1

`speed \: of \: the \: current \: is \to \boxed{4 .5 \: knots }\\ and \: the \\ \: speed \: of \: the \: submarine \: in \: still \: water \: is \: \to \boxed{20.5 \: knots}`

Explanation:

`if \: submarine \: is \: with \: current \: = 25 \: knots \\ if \: submarine \:is \: against \: current \: = 16\: knots \\ let \: the \: submarine \: speed= \: s_(s) \\ let \: the \: currents \: speed= \: s_(c) \\ \boxed{ \underline{hence }}\to \\ s_(s) + s_(c) = 25........(1)\\ s_(s) - s_(c) = 16........(2) \\ you \: can \: now \: make \: a ny\: of \: the \: variables \: \\ the \: subject \: of \: the \: relation....(lets \: choose \: s_(s) \: in \: eq....1 \: ) \\ s_(s) = 25 - s_(c) .....(3 )\: \\ now \: equate \: it \: into \: eq....(2)\\ \\ (25 - s_(c)) -s_(c) = 16 \\ 2s_(c) = 25 - 16 \\ \boxed{s_(c) = (9)/(2 ) = 4.5 \: knots} \\ \\ to \: find \: s_(s) \: we \: \: substitute\: s_(c) \: into \: eq...3\\ s_(s) = 25 - s_(c) = 25 - 4.5 \\ \boxed{s_(s) = 20.5 \: knots}`

♨Rage♨

Answer 2

Let s be the submarine and x be the current.

x + s = 25

x - s = 16

Add the two equations together to get:

2x = 41

Divide both sides by 2:

x = 41/2

x = 20.5

The speed of the submarine is 20.5 knots

The current would be 25 - 20.5 = 4.5 knots.