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For convenience, consider contact with the spring occurs at t = 0, then the integral of the product of the distance x and the x-velocity, xv x dt, over time t is 1 / 2 x 2. The work is the product of the distance times the spring force, which is also dependent on distance; hence the x 2 result.
The area of a regular polygon is half its perimeter multiplied by the distance from its center to its sides, and because the sequence tends to a circle, the corresponding formula–that the area is half the circumference times the radius–namely, A = 1 / 2 × 2πr × r, holds for a circle.
The unit vector ^ has a time-invariant magnitude of unity, so as time varies its tip always lies on a circle of unit radius, with an angle θ the same as the angle of (). If the particle displacement rotates through an angle dθ in time dt , so does u ^ R ( t ) {\displaystyle {\hat {\mathbf {u} }}_{R}(t)} , describing an arc on the unit circle ...
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 curve is a cycloid, and the time is equal to π times the square root of the radius (of the circle which generates the cycloid) over the acceleration of gravity. The tautochrone curve is related to the brachistochrone curve, which is also a cycloid.
where A 1 and A 2 are the centers of the two circles and r 1 and r 2 are their radii. The power of a point arises in the special case that one of the radii is zero. If the two circles are orthogonal, the Darboux product vanishes. If the two circles intersect, then their Darboux product is
At any instant of time, the net force on a body is equal to the body's acceleration multiplied by its mass or, equivalently, the rate at which the body's momentum is changing with time. If two bodies exert forces on each other, these forces have the same magnitude but opposite directions. [1] [2]
In general, if a vector [a 1, a 2, a 3] is represented as the quaternion a 1 i + a 2 j + a 3 k, the cross product of two vectors can be obtained by taking their product as quaternions and deleting the real part of the result. The real part will be the negative of the dot product of the two vectors.