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What if the plane wants to end up directly south? What direction should it steer?

How do you add vectors that don’t have the same (or opposite) direction? Let’s consider adding the following vectors: 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg. SCALE: 1 cm = 5 m.

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What if the plane wants to end up directly south? What direction should it steer?

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  1. How do you add vectors that don’t have the same (or opposite) direction? Let’s consider adding the following vectors: 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg. SCALE: 1 cm = 5 m

  2. Displacement vectors aren’t the only kind of vectors. There are velocity, acceleration, force, and momentum vectors among others. In navigation we use velocity vectors to determine headings when environmental vectors affect our course. Wind vectors affect runners, boats, and especially flying objects. Current vectors will have a profound effect on boats (especially paddle boats) and swimmers.

  3. Relative to the ground, a plane flying against a headwind also has a speed which is the vector sum of the plane’s velocity and the wind’s velocity.

  4. If there is a cross wind, we add the heading vector and the wind vector to find the resultant. Notice that the resultant velocity is faster than the plane’s velocity.

  5. What if the plane wants to end up directly south? What direction should it steer? To do this we need to subtract vectors. Suppose we want A + B = C, but we know C (the resultant) and B (a wind or current vector). If we are trying to solve for A (our heading) we should get it from this equation : A = C - B How do you subtract a vector? Just turn the vector around and add it. But we know the magnitude of A, and the direction of C.

  6. It’s easy to find the additive inverse of B And the magnitude of A We know the direction of C

  7. Start with a large enough C vector, and put the tail of the B inverse vector on the head of the C vector. Now find an angle that will accommodate the magnitude of the A vector. A plane traveling at 100km/hr needs to fly 15˚ east of south in order to fly directly south in a 25km/hr wind from the east. It will do 97 km/hr for the trip.

  8. It is useful to know how to solve vector problems graphically, but it is much faster to solve vector problems algebraically. It is easiest to work with vectors that are normal (perpendicular) to each other. Then we can use Pythagorean’s theorem and basic trigonometry to solve the problem. But any vector can be described as the sum of two orthogonal (normal, or perpendicular) vectors. So let’s learn (review) trig.

  9. For a right triangle a2+b2=c2 sine  = sin  =opposite/hypotenuse cosine  = cos  = adjacent/hypotenuse tangent  = tan  = opposite/adjacent

  10.  = sin-1(opposite/hypotenuse)  = cos-1(adjacent/hypotenuse)  = tan-1(opposite/adjacent) This is how to back-out the angle. Make sure you have degrees highlighted under MODE on your TI-83.

  11. Any vector can be resolved as the sum of two vectors, one parallel to the x-axis and one parallel to the y-axis. Imagine a vector R, at an angle to the x-axis with a magnitude of r.

  12. The magnitude of R is r, the magnitude of Rxis rx, and the magnitude of Ry is ry. So sin = ry/r, cos = rx/r, or rx = rcos ry = rsin. If we are given Rx and Ry, r and  for vector R can be calculated by: r2 = rx2 + ry2  = sin-1(ry/r)  = cos-1(rx/r)  = tan-1(ry/rx)

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