Video Transcript
A small plane can fly at 175
kilometers per hour in still air. The plane flies in a wind that
blows directly out of the west at 36 kilometers per hour. At what angle west of north must
the plane point in order for it to move directly north? How much time is needed for the
plane to reach a point 300 kilometers directly north of its current position?
Let’s call the angle representing
the bearing of the plane 𝜃. We also want to solve for the
amount of time needed for the plane to travel 300 kilometers directly north from
where it currently is. We’ll call this time 𝑡. In this statement, we’re given the
plane’s speed, 175 kilometers per hour in still air. And we’re given the velocity vector
of the wind. We’re told it blows out of the west
at a speed of 36 kilometers per hour. Let’s start on our solution by
drawing a diagram of the plane in travel and this wind affecting its motion.
If we draw in our four compass
directions: north, south, east, and west, we’re told there’s a wind coming out of
the west. We’ve called the speed of that wind
𝑣 sub 𝑤, given as 36 kilometers per hour. We’re told that our plane wants to
be able to fly in a direction due north including the effect of the wind on the
plane’s flight. To do that, the plane will need to
pick its velocity vector. That is, fly in such a direction
that the western component of its velocity vector is counteracted by the eastern
pushing of the wind. When this happens, the plane’s net
motion will be to the north — that is the vector sum of these two velocities. We’re given the magnitude of the
plane’s velocity, its speed which we’ve called 𝑣 sub 𝑝 given as 175 kilometers per
hour.
To solve for 𝜃, we want to
consider the component of the plane’s velocity that will counteract 𝑣 sub 𝑤. We can write this as an
equation. We write that 𝑣 sub 𝑝, the
plane’s speed, multiplied by the sine of the angle 𝜃 is equal to 𝑣 sub 𝑤. This is the condition we enforce so
that the plane’s net motion can be directly north. Dividing both sides by 𝑣 sub 𝑝
and then taking the arcsine of both sides of the equation, we find that 𝜃 is equal
to the inverse sine of the speed of the wind divided by the speed of the plane. When we plug in for those values —
36 kilometers per hour for the wind speed and 175 kilometers per hour for plane
speed — and calculate this expression, we find, to two significant figures, that 𝜃
is 12 degrees. That’s the direction west of north
at which the plane should head so that its net motion is directly north.
Heading this way, we then imagine
that the plane flies a journey of 300 kilometers north. And we want to solve for the time
that this journey would take. To solve for this time, we can
recall that average speed 𝑣 is equal to the distance travelled divided by the time
it takes to travel that distance. We can rearrange this expression so
that it reads 𝑡 is equal to 𝑑 over 𝑣. In our case, 𝑑 is 300 kilometers
and 𝑣 is 𝑣 sub 𝑝, the plane’s speed, multiplied by the cosine of 𝜃. That’s the northerly component of
the plane’s velocity. Plugging in our values for 𝑣 sub
𝑝 and 𝜃, when we calculate this fraction, we find it’s 1.75 hours or to, one
significant figure, 100 minutes. That’s the amount of time that
would take the plane to go 300 kilometers north.
