Example 3.6 from College Physics, 3.5 Addition of Velocities

Adding Velocities: A Boat on a River

A boat is trying to cross a river. Due to the velocity of the river the path traveled by the boat is diagonal. The velocity of the boat, v boat, is equal to zero point seven five meters per second and is in positive y direction. The velocity of the river, v-river, is equal to one point two meters per second and is in positive x direction. The resultant diagonal velocity v total, which makes an angle of theta with the horizontal x axis, is towards north east direction.

Figure 3.46 A boat attempts to travel straight across a river at a speed 0.75 m/s. The current in the river, however, flows at a speed of 1.20 m/s to the right.

Refer to Figure 3.46, which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction of the boat’s velocity relative to an observer on the shore, $v_{tot}$ . The velocity of the boat, $v_{boat}$ , is 0.75 m/s in the $y$ -direction relative to the river and the velocity of the river, $v_{river}$ , is 1.20 m/s to the right.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
We start by choosing a coordinate system with its $x$ -axis parallel to the velocity of the river, as shown in Figure 3.46. Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the $y$ -axis and perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations $v_{tot} = \sqrt{v_{x}^{2} + v_{y}^{2}}$ and $θ = {tan}^{- 1} (v_{y} / v_{x})$ directly.
Solution
The magnitude of the total velocity is

$v_{tot} = \sqrt{v_{x}^{2} + v_{y}^{2}},$
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where

$v_{x} = v_{river} = 1 . 20 m/s$
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and

$v_{y} = v_{boat} = 0 . 750 m/s.$
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Thus,

$v_{tot} = \sqrt{(1 . 20 m/s)^{2} + (0 . 750 m/s)^{2}}$
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yielding

$v_{tot} = 1 . 42 m/s.$
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The direction of the total velocity $θ$ is given by:

$θ = {tan}^{- 1} (v_{y} / v_{x}) = {tan}^{- 1} (0 . 750 / 1 . 20) .$
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This equation gives

$θ = 32 . 0º .$
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Discussion
Both the magnitude $v$ and the direction $θ$ of the total velocity are consistent with Figure 3.46. Note that because the velocity of the river is large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced by the small angle (only $32.0º$ ) the total velocity has relative to the riverbank.

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How did it go?

Strategy

Solution

Discussion

Strategy

Solution

Discussion