Relative Motion in One Dimension — Question 6

Question 6

Infer current and still-water speed from two travel times

A boat travels 600 m downstream in 60 s and returns the same distance upstream in 100 s. Its speed relative to water is the same on both legs, and the water current is steady. Choose downstream as positive.

1. Find the two signed ground velocities.

2. Find the boat speed relative to water and the current velocity.

3. Find average velocity and average speed for the round trip. Why is the latter not the arithmetic mean of the two ground speeds?

Question 6 - Solution

Infer current and still-water speed from two travel times

(a) Attach signs after using distance and time.

$$v_\downarrow=600/60=10\,\mathrm{m/s},\quad v_\uparrow=-600/100=-6\,\mathrm{m/s}$$ Downstream ground velocity is $+600/60=+10$ m/s. Upstream ground velocity is $-600/100=-6$ m/s. The minus sign identifies the direction of the return leg.

(b) Write the two frame equations.

$$u+w=10,\quad u-w=-6\ \Rightarrow\ \boxed{u=2\,\mathrm{m/s},\quad w=8\,\mathrm{m/s}}$$ If u is current speed and w is boat speed through water, $u+w=10$ and $u-w=-6$ in m/s. Adding and subtracting gives u=2 m/s downstream and w=8 m/s. The current helps one leg and opposes the other.

(c) Use the full trip totals.

$$v_{\mathrm{av}}=0/(60+100)=0,\quad\bar v_{\mathrm{speed}}=\frac{1200}{160}=7.5\,\mathrm{m/s}$$ Net displacement is zero, so average velocity is zero. Total distance is 1200 m in 160 s, giving average speed 7.5 m/s. The arithmetic mean $(10+6)/2=8$ m/s would weight unequal-duration legs equally; the correct time weighting gives 7.5.