Question

Solve each problem. See Examples 5 and 6.

Distance and Direction of a Motorboat A motorboat sets out in the direction N 80° 00′ E. The speed of the boat in still water is 20.0 mph. If the current is flowing directly south, and the actual direction of the motorboat is due east, find the speed of the current and the actual speed of the motorboat.

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Accepted Answer

Step 1: Represent the problem using vectors. Let the motorboat's velocity in still water be $$ \vec{v_b} $$ with magnitude 20.0 mph in the direction N 80° E. Express this velocity in terms of its east (x) and north (y) components using trigonometry: $$ v_{bx} = 20.0 	imes \cos(80^\circ) $$ and $$ v_{by} = 20.0 	imes \sin(80^\circ) . $$Step 2: Represent the current's velocity $$ \vec{v_c}  as a $$vector flowing directly south. Since south is the negative y-direction, the current's velocity vector is $$ \vec{v_c} = (0, -v_c) $$, where $$ v_c  is $$the unknown speed of the current. Step 3: The actual velocity of the motorboat $$ \vec{v_r}  is $$the vector sum of the boat's velocity in still water and the current's velocity: $$ \vec{v_r} = \vec{v_b} + \vec{v_c} . $$Given that the actual direction is due east, the north component of $$ \vec{v_r} $$ must be zero. Set up the equation for the y-component: $$ v_{by} - v_c = 0  to $$solve for $$ v_c . $$Step 4: Once $$ v_c  is $$found, calculate the actual speed of the motorboat by finding the magnitude of $$ \vec{v_r} . $$Since the actual velocity is due east, the x-component of $$ \vec{v_r}  is$$ $$ v_{bx} + 0 = v_{bx} . $$The magnitude of $$ \vec{v_r}  is $$then simply $$ v_{bx} . $$Step 5: Summarize the results: the speed of the current is the value found in Step 3, and the actual speed of the motorboat is the magnitude found in Step 4. This completes the problem by combining vector components and using trigonometric relationships.

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Concetti chiave

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Resolving Vectors into Components

Trigonometric Direction Angles