Questions on Algebra: Introduction to vectors, addition and scaling answered by real tutors!

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The heading of a plane is 27.7 degrees NE, and its air speed is 255 mi.h. If the wind is blowing from the south with a velocity of 42.0 mi/h, find the actual direction of travel of the plane, and its ground speed.

The other tutor used the law of cosines and the parallelogram method.
I will use the component method:



The red vector represents the wind's velocity.  Since the wind
is FROM the SOUTH, it is blowing TOWARD the NORTH!  

The green vector represents the plane's velocity.
Note that that the angle 27.7° is measured clockwise from
the vertical (North), so the angle we will use is measured 
counterclockwise from the horizontal (East) calculated as 
90°-27.7° or 62.3°.

         x-components          y-components

Plane     255cos(90°)            255sin(90°)
Wind       42cos(62.3°)           42sin(62.3°)
----------------------------------------------
Sums:     19.52336592            292.1865323

|Resultant| =  = 292.8380635 mi/h

Angle (counterclockwise from East) = tan-1( = 86.177°

To find the bearing clockwise from the North, we subtract from 90° and get 3.823°,

and the bearing is written as N 3.823° E.

I agree with the other tutor on the magnitude of the resultant, 
but we differ just a bit on the angle.

Edwin

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The heading of a plane is 27.7 degrees NE, and its air speed is 255 mi.h. If the wind is blowing from the south with a velocity of 42.0 mi/h, find the actual direction of travel of the plane, and its ground speed.
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ground speed = 292.84 mi/hr
--------------
GS/sin(152.7) = 42/sin(A) A = the angle betwee the heading and the ground track.
-----------
sin(A) = 42*sin(152.7)/GS
A = 3.77 degs
ground track = 27.7 - 3.77 = 23.93 degs, a heading of ~024
--------------
Never seen fractions of degrees used in aviation.
----------------
Update:
Edwin got 3.82 degs, only slightly different from 3.77, not significant.
But, he used the plane's heading as the x-axis, and so should have subtracted the 3.8... from the 27.7 heading to the the direction of the ground track.

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A certain escalator travels at a rate of 10.6 m/min, and its angle of inclination is 32.5 degrees. What is the vertical component of the velocity? How long will it take a passenger to travel 10.0 m vertically?
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vertical component = 10.6*sin(32.5) = 5.64 m/min
----
time to travel 10m = 10/5.64 = 1.7725 minutes = 1min 46.35 seconds
-----------------
Cheers,
Stan H.
============

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if a=2 then 4(2)b and if b=5 then 4(2)(5) 4x2=8 and 8x5=40

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if m=3 then 3+n and if n=2 then 3+2 and 3+2=5

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if x=7 then 7+y and if y=8 then 7+8 and 7+8=15

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37-5(7)
37-35
37-35=2

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Are you really too lazy to type out the words "I don't know"?

You cannot completely evaluate your expression. You have the value of defined, but instead of the value of , you have the value of defined, and is not used in the expression. If you just made a typo, then just replace the letters in your expression with the values given and then do the arithmetic.

John

My calculator said it, I believe it, that settles it

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Thank you for helping your daughter to better understand the world she lives in.
.
"Fysics is fun" ... or is it "Physics is Phun"?
.
Anyhow, in the problem you can ignore friction because it isn't a part of the given information. The way I would start this is to draw a horizontal base line and sketch a rectangle representing the box and using the extended base line as the bottom side of the box. Then put a dot in the center of the box (where the diagonals of the box cross).
.
Label the box 100 kg. Starting at the center dot draw a horizontal vector line to the left and label it 30 newtons. Then return the dot and starting there draw a vertical vector and label it 20 newtons. (Just to keep things sort of to scale, make the length of this vector about two-thirds the length of the first vector of 30 newtons.)
.
Start at the arrowhead point at the end of the 20 newton vector and extend a horizontal line to the left (parallel to the 30 newton vector that you first drew.) Now go to the arrowhead point on the original 30 newton vector and starting there draw a vertical line upward and parallel to the original 20 newton vector. Extend this vertical line until it intersects the horizontal line coming from the arrowhead. You have now formed what is called the force parallelogram, in this case a rectangle having two sides formed by the original 30 newton and 20 newton vectors and the other two sides (meeting at an intersecting point) formed by the vertical line through the left end of the 30 newton vector and the horizontal line through the top of the 20 newton vector arrowhead.
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I hope this is understandable. Now draw a new vector that starts at the center dot of the box and goes left and upward to the intersecting point of the lines from the ends of the two original vectors. Put the arrowhead at the end of this line where it meets the intersecting point. Now we're ready to go to work. This new vector (dot to intersecting point) represents the force (both direction and magnitude) that results from the 30 newton vector and the 20 newton vectors applied to the box.
.
You can now apply the Pythagorean theorem to calculate the magnitude of the force represented by this new vector. The new vector is the hypotenuse of the right triangle with legs of 20 newtons and 30 newtons. By the Pythagorean theorem the hypotenuse (Resultant Force R) is:
.

.

.

.
and the resultant force R is the square root of 1300 or:
.
newtons.
.
This force is equal to the Mass (100 kg) times the Acceleration. So we can write:
.
R = M*A
.
36.0555 = 100 * A
.
Solving for the acceleration we get:
.
A = 36.0555/100
.
which gives an answer of 0.360555 meters/sec^2 for the acceleration in the direction of the Resultant force vector.
.
Instead of the Pythagorean theorem you could have also done this by trigonometry. It's time to use a little trig to find one of the angles that helps to identify the direction of the vector. (I hope your daughter has had some trig so this isn't overwhelming.) You can get the angle between the 30 newton vector and the Resultant force vector by using the arctangent function. For this function the 20 newton side is the opposite side (call it O) and the 30 newton side is the adjacent side (call it A). We write:
.

.

.
You can use a calculator to find that this angle is 33.69 degrees. Again, this is the angle between the 30 newton vector and the Resultant force vector.
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If both forces act together on the box and both are to the left, the resulting combined force is 30 + 20 = 50 newtons. Again using the formula that Force equals Mass times Acceleration we get:
.

.
Divide both sides by 100 and get:
.

.
which reduces to the acceleration A = 0.5 meters/sec^2.
.
Check my math to ensure that I didn't make a dumb mistake somewhere along the line.
.
I hope these answers match yours, and that they may help you to understand the problem a little better.
.

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batty, huh...

you multiply the equation by 2 to clear the fraction (1/2)


FYI ___ it's root, like under trees

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d=sqrt(Gm^1m^2/F)

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Solve the equation to find a formula for r.

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y=5x-2

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...... if , we will have

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The given line r=(-2,3,4)+s(1,1,2) is symmetric form
. The direction vector is thus given by <1,1,2>. A direction vector of the line perpendicular to this is <1,1,-1>. The symmetric equation of the perpendicular equation is thus
, or R = (1,0,2) + t(1,1,-1).

Find the coordinates of the point where the 
line with parametric equations: 

x=1-t, y=3+t and z=3-2t meets the X0Y plane.

Thanks :)

This is when z=0

x=1-t, y=3+t and z=3-2t

Substitute z = 0

                 z=3-2t
                 0=3-2t
                2t=3
                 t=3/2

Substitute t=3/2

x=1-t,     y=3+t      and z=3-2t
x=1-3/2,   y=3+3/2,   and z=3-2(3/2)
x=2/2-3/2, y=6/2+3/2, and z=3-3
x=-1/2,    y=9/2,     and z=0

So the point where the
line x=1-t, y=3+t and z=3-2t
intersects the xy-plane is

(-1/2, 9/2, 0)

Edwin

Hi

 geometric sequence { 2, - 1, 1/2...} r = -(1/2)



Note:  = 2/4 = 1/2



 =  1/8

Hi

4y=2(x+1)^2-1

 y = (1/2)(x+1)^2 - 1/4

the vertex form of a parabola,  where(h,k) is the vertex

 Pt(-1,-1/4) is the vertex

y-intercept, when x  = 0

y = (1/2)*1 - 1/4

y = (1/2) Pt(0,1/4) the y intercept

x-intercept, when y = 0

 0 = (1/2)(x+1)^2 - 1/4

 1/4 = (1/2)(x+1)^2 

  2/4 = x^2 + 2x + 1

   0 = x^2 + 2x + 1/2