2 dx dt v = dv dt d 2x dt 2 a = a = dv dx a = v O P x x The velocity v of the particle is equal to the time derivative of the position coordinate x,dxdtv =and the acceleration a is obtained by differentiating v with respect to t,dvdtd 2xdt 2a =a =orwe can also express a asdvdxa = v
3 dx dt dv dt v = a = d 2x dt 2 dv dx a = v a = O P x x - or + or The velocity v and acceleration a are represented by algebraic numbers which can be positive or negative. A positive value for v indicates that the particle moves in the positive direction, and a negative value that it moves in the negative direction. A positive value for a, however, may mean that the particle is truly accelerated (i.e., moves faster) in the positive direction, or that it is decelerated (i.e., moves more slowly) in the negative direction. A negative value for a is subject to a similar interpretation.
4 x = xo + vt v = vo + at x = xo + vot + at2 v2 = vo + 2a(x - xo ) 2 Two types of motion are frequently encountered: uniform rectilinear motion, in which the velocity v of the particle is constant andx = xo + vtand uniformly accelerated rectilinear motion, in which the acceleration a of the particle is constant andv = vo + at12x = xo + vot at2v2 = vo + 2a(x - xo )2
5 xB = xA + xB/A vB = vA + vB/A aB = aA + aB/A O A B x xA xB/A xB When particles A and B move along the same straight line, the relative motion of B with respect to A can be considered. Denoting by xB/A the relative position coordinate of B with respect to A , we havexB = xA + xB/ADifferentiating twice with respect to t, we obtainvB = vA + vB/AaB = aA + aB/Awhere vB/A and aB/A represent, respectively, the relative velocity and the relative acceleration of B with respect to A.
6 xCxAxBCABWhen several blocks are are connected by inextensible cords, it is possible to write a linear relation between their position coordinates. Similar relations can then be written between their velocities and their accelerations and can be used to analyze their motion.
7 Sometimes it is convenient to use a graphical solution for problems involving rectilinear motion of a particle. The graphical solution most commonly involves x - t, v - t , and a - t curves.At any given time t,av = slope of x - t curvea = slope of v - t curvet1t2tvwhile over any given time interval t1 to t2,v2t2v2 - v1 = a dtòv1t1v2 - v1 = area under a - t curvex2 - x1 = area under v - t curvet1t2txt2x2x2 - x1 = v dtòx1t1t1t2t
8 yThe curvilinear motion of a particle involves particle motion along a curved path. The position P of the particle at a given time is defined by the position vector r joining the origin O of the coordinate system with the point P.vPrsPoOxThe velocity v of the particle is defined by the relationdrdtv =The velocity vector is tangent to the path of the particle, and has a magnitude v equal to the time derivative of the length s of the arc described by the particle:dsdtv =
9 dr dt v = ds dt v = dv dt a = y v P r s Po x O In general, the acceleration a of the particle is not tangent to the path of the particle. It is defined by the relationayPrdvdtsa =PoOx
10 . . . vx = x vy = y vz = z .. .. .. ax = x ay = y az = z y Denoting by x, y, and z the rectangular coordinates of a particle P, the rectangular components of velocity and acceleration of P are equal, respectively, to the first and second derivatives with respect to t of the corresponding coordinates:Pvxvzryjjxikixzzk...yvx = x vy = y vz = zay......ax = x ay = y az = zaxPazThe use of rectangular components is particularly effective in the study of the motion of projectiles.rjkixz
11 rB = rA + rB/A vB = vA + vB/A aB = aA + aB/A y’ For two particles A and B moving in space, we consider the relative motion of B with respect to A , or more precisely, with respect to a moving frame attached to A and in translation with A. Denoting by rB/A the relative position vector of B with respect to A , we haveByrBrB/ArAAx’xzz’rB = rA + rB/ADenoting by vB/A and aB/A , respectively, the relative velocity and the relative acceleration of B with respect to A, we also havevB = vA + vB/AandaB = aA + aB/A
12 yIt is sometimes convenient to resolve the velocity and acceleration of a particle P into components other than the rectangular x, y, and z components. For a particle P moving along a path confined to a plane, we attach to P the unit vectors et tangent to the path and en normal to the path and directed toward the center of curvature of the path.Cv 2ran = endvdtat = etPxOThe velocity and acceleration are expressed in terms of tangential and normal components. The velocity of the particle isv = vetThe acceleration isdvdtv2ra = et en
13 v = vet v2 dv a = et + en r dt y C v 2 r an = en dv at = et dt P x O In these equations, v is the speed of the particle and r is the radius of curvature of its path. The velocity vector v is directed along the tangent to the path. The acceleration vector a consists of a component at directed along the tangent to the path and a component an directed toward the center of curvature of the path,
14 a = (r - rq2)er + (rq + 2rq)eq When the position of a particle moving in a plane is defined by its polar coordinates r and q, it is convenient to use radial and transverse components directed, respectively, along the position vector r of the particle and in the direction obtained by rotating r through 90o counterclockwise. Uniterr = r erPqxOvectors er and eq are attached to P and are directed in the radial and transverse directions. The velocity and acceleration of the particle in terms of radial and transverse components is..v = rer + rqeq.......a = (r - rq2)er + (rq + 2rq)eq
15 a = (r - rq2)er + (rq + 2rq)eq ..erv = rer + rqeq.......r = r erPa = (r - rq2)er + (rq + 2rq)eqqxOIn these equations the dots represent differentiation with respect to time. The scalar components of of the velocity and acceleration in the radial and transverse directions are therefore..vr = r vq = rq.......ar = r - rq aq = rq + 2rqIt is important to note that ar is not equal to the time derivative of vr, and that aq is not equal to the time derivative of vq.