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v tot x = v w x size 12{v rSub { size 8{"tot"x} } =v rSub { size 8{wx} } } {}

and

v tot y = v w y + v p . size 12{v rSub { size 8{"tot"y} } =v rSub { size 8{wx} } +v rSub { size 8{p} } "."} {}

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

v w y = v tot x = v tot cos 110º . size 12{v rSub { size 8{wx} } =v rSub { size 8{"tot"x} } =v rSub { size 8{"tot"} } "cos110" rSup { size 8{o} } "."} {}

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

v w y = ( 38.0 m/s ) ( –0.342 ) = –13 m/s.

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

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

v tot y = v w y + v p size 12{v rSub { size 8{"tot"y} } =v rSub { size 8{wx} } +v rSub { size 8{p} } } {}

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

v w y = ( 38 . 0 m/s ) ( 0 . 940 ) 45 . 0 m/s = 9 . 29 m/s. size 12{v rSub { size 8{wy} } = \( "38" "." 0" m/s" \) \( 0 "." "940" \) - "45" "." 0" m/s"= - 9 "." "29"" m/s."} {}

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

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

v w = v w x 2 + v w y 2 = ( 13 . 0 m/s ) 2 + ( 9 . 29 m/s ) 2

so that

v w = 16 . 0 m/s . size 12{v rSub { size 8{w} } ="16" "." 0" m/s."} {}

The direction is:

θ = tan 1 ( v w y / v w x ) = tan 1 ( 9 . 29 / 13 . 0 ) size 12{θ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{wy} } /v rSub { size 8{wx} } \) ="tan" rSup { size 8{ - 1} } \( - 9 "." "29"/ - "13" "." 0 \) } {}

giving

θ = 35 . . size 12{θ="35" "." 6º"."} {}

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 [link] . 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.

Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity.

Summary

  • Velocities in two dimensions are added using the same analytical vector techniques, which are rewritten as
    v x = v cos θ size 12{v rSub { size 8{x} } =v"cos"θ} {}
    v y = v sin θ size 12{v rSub { size 8{y} } =v"sin"θ} {}
    v = v x 2 + v y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {}
    θ = tan 1 ( v y / v x ) . size 12{θ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) } {}
  • Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame.
  • Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move relative to one another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s).

Conceptual questions

If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall straight down as viewed by a person standing at the side of the road? Under what condition would this occur? How would the motion of the ball appear to the person who threw it?

The hat of a jogger running at constant velocity falls off the back of his head. Draw a sketch showing the path of the hat in the jogger’s frame of reference. Draw its path as viewed by a stationary observer.

Problems&Exercises

Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45º size 12{"45º"} {} south of east. What was his total displacement? (b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air? (c) What was his total displacement relative to the air mass?

(a) 35 . 8 km size 12{"35" "." 8" km"} {} , 45º size 12{"45º"} {} south of east

(b) 5 . 53 m/s size 12{5 "." "53 m/s"} {} , 45º size 12{"45º"} {} south of east

(c) 56 . 1 km size 12{"56" "." 1" km"} {} , 45º size 12{"45"º} {} south of east

Practice Key Terms 5

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Source:  OpenStax, 2d kinematics. OpenStax CNX. Sep 04, 2015 Download for free at http://legacy.cnx.org/content/col11879/1.3
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