Relative Motion in One Dimension — Question 1

Question 1

Walking west in a train can still mean moving east on the ground

A train moves east at 20 m/s relative to the ground. A passenger walks west at 2 m/s relative to the train. Use one common east-positive axis and a shared time coordinate. For collinear motion, $v_{A/G}=v_{A/B}+v_{B/G}$; each subscript names the object and its reference frame.

1. Find the passenger’s ground velocity and explain its direction.

2. Find the passenger’s displacement in each frame during 5 s of steady walking.

3. What westward walking speed relative to the train would make the passenger stationary relative to the ground?

Question 1 - Solution

Walking west in a train can still mean moving east on the ground

(a) Keep the westward sign.

$$\boxed{v_{P/G}=v_{P/T}+v_{T/G}=-2+20=18\,\mathrm{m/s}}$$ $v_{P/T}=-2$ m/s and $v_{T/G}=+20$ m/s. Thus $v_{P/G}=-2+20=+18$ m/s, eastward. The word west refers to motion relative to the train, not necessarily relative to the ground.

(b) Use the velocity belonging to each frame.

$$\Delta x_{P/T}=-2(5)=-10\,\mathrm m,\quad\Delta x_{P/G}=18(5)=90\,\mathrm m=-10+100\,\mathrm m$$ Over 5 s the passenger moves $-10$ m in train coordinates but $+90$ m in ground coordinates. The train itself moves $+100$ m, and $90=-10+100$. Displacements over a nonzero interval depend on the translating frame.

(c) Solve for cancellation in the ground frame.

$$0=v_{P/T}+20\ \Rightarrow\ v_{P/T}=-20\,\mathrm{m/s}$$ Stationary ground motion requires $v_{P/T}+20=0$, hence $v_{P/T}=-20$ m/s. This is a kinematic requirement, not a claim that such a walking speed is practical. The required direction is west relative to the train.