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Equation [3] involves the average velocity v + v 0 / 2 . Intuitively, the velocity increases linearly, so the average velocity multiplied by time is the distance traveled while increasing the velocity from v 0 to v, as can be illustrated graphically by plotting velocity against time as a straight line graph. Algebraically, it follows ...
The linear motion can be of two types: uniform linear motion, with constant velocity (zero acceleration); and non-uniform linear motion, with variable velocity (non-zero acceleration). The motion of a particle (a point-like object) along a line can be described by its position x {\displaystyle x} , which varies with t {\displaystyle t} (time).
A twist is a screw used to represent the velocity of a rigid body as an angular velocity around an axis and a linear velocity along this axis. All points in the body have the same component of the velocity along the axis, however the greater the distance from the axis the greater the velocity in the plane perpendicular to this axis.
Angular velocity: the angular velocity ω is the rate at which the angular position θ changes with respect to time t: = The angular velocity is represented in Figure 1 by a vector Ω pointing along the axis of rotation with magnitude ω and sense determined by the direction of rotation as given by the right-hand rule.
Since the angular velocity ω = v/r is constant, the area swept out in a time Δt equals ω r 2 Δt; hence, equal areas are swept out in equal times Δt. In uniform linear motion (i.e., motion in the absence of a force, by Newton's first law of motion), the particle moves with constant velocity, that is, with constant speed v along a line.
The linear velocity of a rigid body is a vector quantity, equal to the time rate of change of its linear position. Thus, it is the velocity of a reference point fixed to the body. During purely translational motion (motion with no rotation), all points on a rigid body move with the same velocity. However, when motion involves rotation, the ...
Translational symmetry of the problem results in the center of mass = = = moving with constant velocity, so that C = L 0 t + C 0, where L 0 is the linear velocity and C 0 is the initial position. The constants of motion L 0 and C 0 represent six integrals of the motion.
The Stokes drift velocity ū S, which is the particle drift after one wave cycle divided by the period, can be estimated using the results of linear theory: [38] u ¯ S = 1 2 σ k a 2 cosh 2 k ( z + h ) sinh 2 k h e k , {\displaystyle {\bar {\mathbf {u} }}_{S}={\tfrac {1}{2}}\sigma ka^{2}{\frac {\cosh 2k(z+h)}{\sinh ^{2}kh}}\mathbf {e ...