Figure 4: 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. What is the total displacement of the boat relative to the shore?

Refer to Figure 4, 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, vtotvtot size 12{v rSub { size 8{"tot"} } } {}. The velocity of the boat, vboatvboat size 12{v rSub { size 8{"boat"} } } {}, is 0.75 m/s in the y y size 12{y} {} -direction relative to the river and the velocity of the river, vrivervriver size 12{v rSub { size 8{"river"} } } {}, is 1.20 m/s to the right.

Strategy

We start by choosing a coordinate system with its xx-axis parallel to the velocity of the river, as shown in Figure 4. Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the yy-axis and perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations vtot=vx2+vy2vtot=vx2+vy2 size 12{v rSub { size 8{"tot"} } = sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {} and θ=tan−1(vy/vx)θ=tan−1(vy/vx) size 12{θ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) } {} directly.

Solution

The magnitude of the total velocity is

where

and

Thus,

yielding

The direction of the total velocity θ θ size 12{θ} {} is given by:

This equation gives

Discussion

Both the magnitude vv and the direction θθ of the total velocity are consistent with Figure 4. 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 32.0º32.0º size 12{"32.0º"} {}) the total velocity has relative to the riverbank.

Calculate the wind velocity for the situation shown in Figure 5. The plane is known to be moving at 45.0 m/s due north relative to the air mass, while its velocity relative to the ground (its total velocity) is 38.0 m/s in a direction 20.0º20.0º size 12{"20" "." 0 rSup { size 8{o} } } {} west of north.

Figure 5: An airplane is known to be heading north at 45.0 m/s, though its velocity relative to the ground is 38.0 m/s at an angle west of north. What is the speed and direction of the wind?

Strategy

In this problem, somewhat different from the previous example, we know the total velocity vtotvtot size 12{v rSub { size 8{ bold "tot"} } } {} and that it is the sum of two other velocities, vwvw size 12{v rSub { size 8{w} } } {} (the wind) and vpvp size 12{v rSub { size 8{p} } } {} (the plane relative to the air mass). The quantity vpvp size 12{v rSub { size 8{p} } } {} is known, and we are asked to find vwvw size 12{v rSub { size 8{w} } } {}. None of the velocities are perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we can find the components of vwvw size 12{v rSub { size 8{w} } } {}, then we can combine them to solve for its magnitude and direction. As shown in Figure 5, we choose a coordinate system with its x-axis due east and its y-axis due north (parallel to vpvp size 12{v rSub { size 8{p} } } {}). (You may wish to look back at the discussion of the addition of vectors using perpendicular components in Vector Addition and Subtraction: Analytical Methods.)

Solution

Because vtotvtot size 12{v rSub { size 8{ bold "tot"} } } {} is the vector sum of the vwvw and vpvp, its x- and y-components are the sums of the x- and y-components of the wind and plane velocities. Note that the plane only has vertical component of velocity so vpx=0vpx=0 and vpy=vpvpy=vp. That is,

and

We can use the first of these two equations to find v wx v wx size 12{v rSub { size 8{ ital "wx"} } } {} :

Because v tot = 38 . 0 m / s v tot = 38 . 0 m / s size 12{v rSub { size 8{ ital "tot"} } ="38" "." 0m/s} {} and cos 110º = – 0.342 cos 110º = – 0.342 size 12{"cos""110º""=""-""0.342"} {} we have

The minus sign indicates motion west which is consistent with the diagram.

Now, to find v w y v w y size 12{v rSub { size 8{ ital "wy"} } } {} we note that

Here vtoty= vtotsin 110ºvtoty= vtotsin 110º size 12{v rSub { size 8{"tot"y} } =v rSub { size 8{"tot"} }  = v rSub { size 8{"tot"} }  "sin 110º"} {}; thus,

This minus sign indicates motion south which is consistent with the diagram.

Now that the perpendicular components of the wind velocity vwxvwx size 12{v rSub { size 8{wx} } } {} and vwyvwy size 12{v rSub { size 8{wy} } } {} are known, we can find the magnitude and direction of vwvw size 12{v rSub { size 8{w} } } {}. First, the magnitude is

so that

The direction is:

giving

Discussion

The wind’s speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in Figure 5. Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total velocity significantly less than its velocity relative to the air mass as well as heading in a different direction.

An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth?

Figure 7: The motion of a coin dropped inside an airplane as viewed by two different observers. (a) An observer in the plane sees the coin fall straight down. (b) An observer on the ground sees the coin move almost horizontally.

Strategy

Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes.

Solution for (a)

Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found using the equation:

Substituting known values into the equation, we get

yielding

We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane.

Solution for (b)

Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity for the coin relative to the ground is vy=−5.42m/s vy=−5.42m/s , the same as found in part (a). In contrast to part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and vx =260 m/s vx =260 m/s size 12{"v subx =260 m/s"} {} . The x- and y-components of velocity can be combined to find the magnitude of the final velocity:

Thus,

yielding

The direction is given by:

so that

Discussion

In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do not add like ordinary numbers—the final velocity v in part (b) is not (260 – 5.42) m/s(260 – 5.42) m/s size 12{ \( "260 – 5" "." "42" \) " m/s"} {}; rather, it is 260.06 m/s260.06 m/s size 12{"260" "." "06 m/s"} {}. The velocity’s magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the two observers see very different paths. (See Figure 7.) In addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path.