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A boy swims 1 km upstream and returns downstream. Compare his total swimming time in still water versus water with a steady speed of 0.5 m/s.

Physics

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Given:
Boy's speed in still water, vb=1.2 m/sv_b = 1.2 \text{ m/s}
Distance to swim upstream (one way), d=1 km=1000 md = 1 \text{ km} = 1000 \text{ m}

1. When the water is still

  1. Determine the total distance.

    The boy swims 1 km1 \text{ km} upstream and returns 1 km1 \text{ km} downstream.
    Total distance D=1000 m+1000 m=2000 mD = 1000 \text{ m} + 1000 \text{ m} = 2000 \text{ m}.

  2. Calculate the total time.

    When the water is still, the boy's speed is simply his speed in still water.

    T1=DvbT_1 = \frac{D}{v_b} T1=2000 m1.2 m/sT_1 = \frac{2000 \text{ m}}{1.2 \text{ m/s}} T1=2000012 sT_1 = \frac{20000}{12} \text{ s} T1=50003 sT_1 = \frac{5000}{3} \text{ s} T1≈1666.67 sT_1 \approx 1666.67 \text{ s}

    The total time when the water is still is 1666.67 s\boxed{1666.67 \text{ s}}.

    2. When the water has a steady speed of 0.5 m/s0.5 \text{ m/s}
    Given water speed, vw=0.5 m/sv_w = 0.5 \text{ m/s}.

  3. Calculate the speed when swimming upstream.

    When swimming upstream, the water current opposes the boy's motion.

    vup=vb−vwv_{up} = v_b - v_w vup=1.2 m/s−0.5 m/sv_{up} = 1.2 \text{ m/s} - 0.5 \text{ m/s} vup=0.7 m/sv_{up} = 0.7 \text{ m/s}

  4. Calculate the time taken to swim upstream.

    tup=dvupt_{up} = \frac{d}{v_{up}} tup=1000 m0.7 m/st_{up} = \frac{1000 \text{ m}}{0.7 \text{ m/s}} tup=100007 st_{up} = \frac{10000}{7} \text{ s}

  5. Calculate the speed when swimming downstream.

    When swimming downstream, the water current aids the boy's motion.

    vdown=vb+vwv_{down} = v_b + v_w vdown=1.2 m/s+0.5 m/sv_{down} = 1.2 \text{ m/s} + 0.5 \text{ m/s} vdown=1.7 m/sv_{down} = 1.7 \text{ m/s}

  6. Calculate the time taken to swim downstream.

    tdown=dvdownt_{down} = \frac{d}{v_{down}} tdown=1000 m1.7 m/st_{down} = \frac{1000 \text{ m}}{1.7 \text{ m/s}} tdown=1000017 st_{down} = \frac{10000}{17} \text{ s}

  7. Calculate the total time.

    T2=tup+tdownT_2 = t_{up} + t_{down} T2=100007 s+1000017 sT_2 = \frac{10000}{7} \text{ s} + \frac{10000}{17} \text{ s} T2=10000(17+117) sT_2 = 10000 \left(\frac{1}{7} + \frac{1}{17}\right) \text{ s} T2=10000(17+77×17) sT_2 = 10000 \left(\frac{17 + 7}{7 \times 17}\right) \text{ s} T2=10000(24119) sT_2 = 10000 \left(\frac{24}{119}\right) \text{ s} T2=240000119 sT_2 = \frac{240000}{119} \text{ s} T2≈2016.81 sT_2 \approx 2016.81 \text{ s}

    The total time when the water has a steady speed of 0.5 m/s0.5 \text{ m/s} is 2016.81 s\boxed{2016.81 \text{ s}}.

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