Thus, if we refer the position, velocity, and acceleration vectors to a fixed cartesian coordinate syst have r(t)=r(t)i+y(t)j+2(t)k v(t)=tr(t1)i+vy(t)j+v(t)k=x(t)i+y(t)j+i(t)k=r(t) a(t)=ar(t)i+ay(t)j+a2(t)k=ix(t)i+iy(t)j+i2(t)k=v(t) (5) Here, the speed is given by=V吗+吗+, and the magnitude of the acceleration is a=V"+吗+吗 The advantages of cartesian coordinate systems is that they are simple to use, and that if a is constant, or a function of time only, we can integrate each component of the acceleration and velocity independently as hown in the ballistic motion example Example Circular Motion We consider motion of a particle along a circle of radius R at a constant speed vo. The parametrization of circle in terms of the arc length is r(s)=RcosRi+Rsin(R)j Since we have a constant speed vo, we have s= vot. Thus, r(t)=Rcos(p-i+ Rsin(p )3 The velocity 2=-(m+mc方Thus, if we refer the position, velocity, and acceleration vectors to a fixed cartesian coordinate system, we have, r(t) = x(t)i + y(t)j + z(t)k (3) v(t) = vx(t)i + vy(t)j + vz(t)k = ˙x(t)i + ˙y(t)j + ˙z(t)k = r˙(t) (4) a(t) = ax(t)i + ay(t)j + az(t)k = ˙vx(t)i + ˙vy(t)j + ˙vz(t)k = v˙(t) (5) Here, the speed is given by v = q v 2 x + v 2 y + v 2 z , and the magnitude of the acceleration is a = q a 2 x + a 2 y + a 2 z . The advantages of cartesian coordinate systems is that they are simple to use, and that if a is constant, or a function of time only, we can integrate each component of the acceleration and velocity independently as shown in the ballistic motion example. Example Circular Motion We consider motion of a particle along a circle of radius R at a constant speed v0. The parametrization of a circle in terms of the arc length is r(s) = R cos( s R )i + R sin( s R )j . Since we have a constant speed v0, we have s = v0t. Thus, r(t) = R cos(v0t R )i + R sin(v0t R )j . The velocity is v(t) = dr(t) dt = −v0 sin(v0t R )i + v0 cos(v0t R )j , 5