## Position Velocity and Acceleration

Position of a particle P moving in space is identified by a vector connecting the particle P with the origin O of reference frame E. This vector is called the position vector, and it is denoted by the symbol r. Because the position of the terminal point of r depends on time, r is a vector function of time t: r = r(t). The vector r can be written as r = (xi ei + x2 e2 + x3 e3) (2.7a)

in which xi, x2, and x3 are all functions of time t:

The velocity v and the acceleration a of particle P in the reference frame E are then defined as follows:

v = dr/dt = (dx1/dt e1 + dx2/dt e2 + dx3/dt e3) (2.8a)

a = dv/dt = (dv1/dt e1 + dv2/dt e2 + dv3/dt e3) (2.8b)

These equations indicate that to compute the velocity and acceleration of a point P with respect to a reference frame fixed on earth, we need to express the position vector of P in terms of the unit vectors fixed on earth and then take the time derivatives of vector projections on the coordinate axes. This procedure is illustrated next with an example.

Example 2.3. Particle Path, Velocity, and Acceleration. The position vector connecting a fixed point O in the reference frame E to a moving point P in space is given by the expression:

Determine the velocity and acceleration of point P in reference frame E.

Solution: The unit vectors e1, e2 are constants in E so their time derivatives will be zero. Using differentiation by parts, we find v = dr/ dt

= 6t [cos (2t2) e1 + sin (2t2) e2] + (1.67 + 3 t2) (4t) [-sin (2t2) e1 + cos (2t2) e2]

= [6t cos (2t2) - (6.7t + 12t3) sin (2t2)] e1 + [6t sin (2t2) + (6.7t + 12t3) cos(2t2)] e2

= [6cos (2t2) - 24t2sin (2t2) - (6.7 + 36t2)sin (2t2) + (26.8t2 + 48t4) cos (2t2)] e1

+ [6sin (2t2) + 24t2cos (2t2) + (6.7 + 36t2) cos (2t2) - (26.8t2 + 48t4) sin (2t2)] e2

Note that in this book we typically refer to velocity (acceleration) with respect to a reference frame fixed on earth as simply velocity (acceleration).

## Getting Started With Dumbbells

The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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