College Physics

Example 3.6 Adding Velocities: A Boat on a River

Figure 3.47A 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. What is the

total displacement of the boat relative to the shore?

Refer toFigure 3.47, 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,vtot. The velocity of the boat,vboat, is 0.75 m/s in they-direction relative to the river and the velocity of

the river,vriver, is 1.20 m/s to the right.

Strategy

We start by choosing a coordinate system with itsx-axis parallel to the velocity of the river, as shown inFigure 3.47. Because the boat is

directed straight toward the other shore, its velocity relative to the water is parallel to they-axis and perpendicular to the velocity of the river.

Thus, we can add the two velocities by using the equationsvtot= v^2 x+vy^2 andθ= tan−1(vy/vx)directly.

Solution

The magnitude of the total velocity is

(3.76)

vtot= vx^2 +vy^2 ,

where

vx=vriver= 1.20 m/s (3.77)

and

vy=vboat= 0.750 m/s. (3.78)

Thus,

(3.79)

vtot= (1.20 m/s)^2 + (0.750 m/s)^2

yielding

vtot= 1.42 m/s. (3.80)

The direction of the total velocityθis given by:

θ= tan−1(v (3.81)

y/vx) = tan

−1(0.750 / 1.20).

This equation gives

θ= 32.0º. (3.82)

Discussion

Both the magnitudevand the directionθof the total velocity are consistent withFigure 3.47. 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 (only32.0º) the total

velocity has relative to the riverbank.

110 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS

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