Acceleration — Question 8

Question 8

An accelerating reference contributes its own acceleration

A cart moves along a platform parallel to a ground-fixed east-positive line. The platform origin has ground position $X_O(t)=t^2$, while the cart has platform-relative coordinate $x'(t)=8t-2t^2$. Both axes point east. In these expressions $t$ is numerical seconds and positions are numerical metres, over $0\leq t\leq3$.

At every event the cart’s ground position satisfies $$X(t)=X_O(t)+x'(t).$$ Use $v=dx/dt$ and $a=dv/dt$ in the appropriate reference. This is a coordinate comparison; no force law is required.

1. Find the cart’s ground position function, its ground velocity and acceleration, and its platform-relative velocity and acceleration.

2. Derive the acceleration relation connecting the two references and the platform origin. Check it numerically.

3. At 1 s and 3 s, compare the cart’s motion direction and whether it is speeding up or slowing down in each reference. Use the smooth polynomial extensions at the endpoint if needed.

4. A student claims that acceleration is unchanged by any moving origin because a change of origin cancels in differences. Identify the missing distinction. Would a fixed origin offset or an origin moving with constant velocity contribute acceleration?

Question 8 - Solution

An accelerating reference contributes its own acceleration

(a) Combine positions before describing the ground motion.

$$\boxed{X(t)=8t-t^2\ \mathrm{m}},\qquad \boxed{v_X(t)=8-2t\ \mathrm{m/s}},\qquad \boxed{a_X=-2\ \mathrm{m/s^2}}.$$ The numerical-time convention applies to these formulas. Relative to the platform, $$\boxed{v'(t)=8-4t\ \mathrm{m/s}},\qquad \boxed{a'=-4\ \mathrm{m/s^2}}.$$

(b) Differentiate the same-event coordinate relation twice.

The first derivative gives $v_X=v_O+v'$. The second gives $$\boxed{a_X=a_O+a'}.$$ Here $X_O=t^2$ implies $a_O=+2\ \mathrm{m/s^2}$, so $-2=+2+(-4)$ in $\mathrm{m/s^2}$. The nonzero acceleration of the origin supplies the difference between the cart’s two reported accelerations.

(c) Compare velocity with acceleration in each reference.

$$\begin{aligned}t=1:&\quad v_Xa_X=6(-2)<0,\quad v\!\,^{\prime}a'=4(-4)<0,\\t=3:&\quad v_Xa_X=2(-2)<0,\quad v\!\,^{\prime}a'=(-4)(-4)>0.\end{aligned}$$

At 1 s, the ground velocity is $+6\ \mathrm{m/s}$ and the local velocity is $+4\ \mathrm{m/s}$. Both descriptions are eastward with negative acceleration, hence decreasing speed.

At 3 s, the ground velocity is $+2\ \mathrm{m/s}$, so the cart still moves east and slows relative to the ground. The local velocity is $-4\ \mathrm{m/s}$; together with local acceleration $-4\ \mathrm{m/s^2}$, this means westward motion with increasing speed relative to the platform. These statements describe different reference histories and are compatible.

(d) A time-dependent origin is not a fixed offset.

A fixed spatial offset has zero first and second derivatives. An origin moving with constant velocity has a nonzero first derivative but zero second derivative, so it contributes no acceleration. An accelerating origin has a nonzero second derivative and cannot be removed as a constant. The student’s cancellation argument applies only to the appropriate time-independent or constant-velocity transformations, not to every moving reference.