By itself, the specified vector does nothing in terms of the mathematical description of motion, so it is considered in projections on the coordinate axes. This may be one coordinate axis (the beam), two (plane) or three (space). To find the projection, you need to drop perpendiculars from the ends of the vector on the axis.
Projection is like a "shadow" of the vector. If the body is moving perpendicular to the considered axis, the projection degenerate to the point and will have a value of zero. When moving parallel to a coordinate axis, the projection coincides with the module of the vector. And when the body moves so that its velocity vector is directed at some angle φ to the x-axis and the projection on the x-axis is cut: V(x)=V•cos(φ), where V is the module of the velocity vector. Projection is positive when the direction of the velocity vector coincides with the positive direction of the coordinate axis, and negative otherwise.
Let the motion of a point set to the coordinate equations: x=x(t), y=y(t), z=z(t). Then a function of speed, projected onto the three axes, will have the form, respectively, V(x)=dx/dt=x'(t), V(y)=dy/dt=y'(t), V(z)=dz/dt=z'(t), that is, to find the speed you need to take derivatives. The velocity vector will be expressed by the equation V=V(x)•i+V(y)•j+V(z)•k, where i, j, k are unit vectors of coordinate axes x, y, z. The module speed can be calculated by the formula V=√(V(x)^2+V(y)^2+V(z)^2).
Through the guides of the cosines of the velocity vector and unit segments of the axes you can set the direction of the vector, discarding his module. For a point that moves in a plane only two coordinates, x and y. If a body makes a circular motion, the direction of the velocity vector is constantly changing, and the module may remain constant or vary over time.
Advice 2: How to find the projection of on the axis
To find the projection of the vector or line segment on the coordinate axis, you need to drop perpendiculars to the extreme points on each axis. If you know the coordinates of the vector or segment, its projection on the axis can be calculated. The same can be done if you know the length of the vector and the angle between it and the axle.
You will need
- - the concept of the Cartesian coordinate system;
- - trigonometric functions;
- - action vectors.
Draw a vector or a line segment in the coordinate system. Then, from one end of the segment or vector drop perpendiculars to each of the axes. At the intersection of the perpendicular and each axis mark the point. Repeat this procedure for the second end of the segment or vector.
Measure the distance from the origin to each point of intersection with the coordinate system. On each axis from a greater distance, subtract the least - this will be the cut or projection of the vector on each of the axles.
If you know the coordinates of the ends of the vector or line segment to find its projection on the axis, from the coordinates of the end subtract the corresponding coordinates of the beginning. If the value is negative, take its module. The minus sign indicates that the projection is in the negative part of the coordinate axis. For example, if the coordinates of beginning of vector (-2;4;0) and the coordinates of the end (2;6;4), the projection on the axis OX is equal to 2-(-2)=4, on the OY axis: 6-4=2 on axis OZ: 4-0=4.
If the coordinates of the vector, then they are the projections on the respective axis. For example, if the vector has coordinates (4;-2;5), it means that the projection on the axis OX equal to 4, on the OY axis: 2 axis OZ: 5. If the coordinate vector is 0, and its projection on this axis is also equal to 0.
In that case, if you know the length of the vector and the angle between it and the axis (in polar coordinates), in order to find its projection on this axis, you need to multiply the length of this vector on tothe axis of theNUS of the angle between the axis and the vector. For example, if it is known that the vector length is 4 cm and the angle between him and the OX axis in the coordinate system XOY is equal to 60º.
To find its projection on the OX axis, multiply by 4 cos(60º). Calculation 4•cos(60º)=4•1/2=2 cm, Find the projection on the OY axis by finding the angle between it and the vector 90º-60º=30º. Then its projection on this axis will be 4•cos(30º)=4•0,866=3,46 cm