Relative Motion in One Dimension — Question 7

Question 7

A moving walkway changes round-trip timing

A 60 m long moving walkway runs east at 1 m/s. A person walks east along it and then west back to the original ground endpoint, maintaining a speed of 2 m/s relative to the walkway in either direction. The walkway is long enough to support both legs.

1. Find ground velocities and travel times for the two legs.

2. Find round-trip average ground speed and average ground velocity.

3. Find the total walking distance relative to the moving belt and explain why it differs from the ground distance.

Question 7 - Solution

A moving walkway changes round-trip timing

(a) Add the belt velocity with the proper walking sign.

$$v_E=2+1=3,\quad v_W=-2+1=-1\quad\mathrm{m/s};\qquad t_E=60/3=20,\quad t_W=60/1=60\quad\mathrm s$$ Eastward ground velocity is $2+1=3$ m/s, so that leg takes 20 s. Westward ground velocity is $-2+1=-1$ m/s, so the return takes 60 s. Walking speed through the belt is equal, but ground speeds are not.

(b) Use the ground endpoints and elapsed time.

$$\boxed{\bar v_{\mathrm{speed}}=\frac{60+60}{20+60}=1.5\,\mathrm{m/s}},\quad v_{\mathrm{av}}=0$$ The round trip lasts 80 s and covers 120 m relative to the ground. Average ground speed is 1.5 m/s and average ground velocity is zero. Opposite displacements cancel, whereas traveled lengths add.

(c) Measure distance in the named reference frame.

$$s_{P/B}=\int_0^{80}|v_{P/B}|\,dt=2(80)=160\,\mathrm m\ne s_{P/G}=120\,\mathrm m$$ Relative to the belt, the walking distance is $2(20+60)=160$ m. This exceeds the 120 m ground distance because the belt continues moving during the longer westward return. Distance traveled over time is frame-dependent; the ground endpoint separation at one instant is a different quantity.