= Inertial motion in an accelerating frame
{title2=$\mathbf a'=\mathbf a-\ddot{\mathbf R}$}
Constant <velocity> depends on the <reference frame>. For nonrotating axes and a moving origin of laboratory <position> $\mathbf R(t)$, subtracting that origin gives
$$
\mathbf r'=\mathbf r-\mathbf R(t),\qquad
\mathbf v'=\mathbf v-\dot{\mathbf R}(t),\qquad
\mathbf a'=\mathbf a-\ddot{\mathbf R}(t).
$$
If $\mathbf R(t)=\mathbf R_0+\mathbf U t$ with constant <velocity> $\mathbf U$, the <Galilean transformation> leaves <acceleration> unchanged. The cart moving at $1.5\,\mathrm{m/s}$ in the laboratory is at rest in a frame moving right at $1.5\,\mathrm{m/s}$. Both descriptions obey <Newton's first law>.
Instead choose an origin with $R_x(t)=\tfrac12At^2$, where the stipulated constant $A=0.50\,\mathrm{m/s^2}$ points right. The same cart has
$$
x'(t)=(1.5\,\mathrm{m/s})t-\tfrac12(0.50\,\mathrm{m/s^2})t^2,\qquad
v_x'(t)=1.5\,\mathrm{m/s}-(0.50\,\mathrm{m/s^2})t,\qquad
a_x'=-0.50\,\mathrm{m/s^2}.
$$
At $t=3\,\mathrm s$, the cart is instantaneously at rest relative to this accelerating origin, but its relative <acceleration> is nonzero. In this <translating accelerated reference frame>, <Newton's second law> must include the effective inertial <force> $-mA=-1.0\,\mathrm N$. This term accounts for the accelerating coordinates; the actual <external forces> on the cart still sum to zero. No rotation-dependent terms are present because the axes remain parallel to the laboratory axes.
Back to article page
