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Two boats start together and race across a 68-km-wide lake and back. Boat A goes across at 68 km/h and returns at 68 km/h. Boat B goes across at 34 km/h, and its crew, realizing how far behind it is getting, returns at 102 km/h. Turnaround times are negligible, and the boat that completes the round trip first wins.

a. Which boat wins and by how much?
b. What is the average velocity of the winning boat?

Answer :

okpalawalter8

Answer:

a

The  winning boat is  boat A

b

[tex]v_{avg} =0 m/s [/tex]

Explanation:

From the question we are told that

  The  width of the lake is  [tex]w =  68 \  km  [/tex]

   The  speed of boat A to and fro  is  [tex]v_a  =  68km/h[/tex]

   The speed of boat B going is  [tex]v_b  =  34 km/h[/tex]

    The speed of boat B coming back is  [tex]v_B =  102 \ km/h[/tex]

Generally the time taken by boat A is mathematically represented as

    [tex]t_A  =   \frac{w}{ v_a}  + \frac{w}{ v_a}[/tex]

      [tex]t_A  =  \frac{68}{68}  +  \frac{68}{68}[/tex]

     [tex]t_A  =  2 \  hours[/tex]

Generally the time taken by boat B is mathematically represented as

      [tex]t_B  =   \frac{w}{ v_b}  + \frac{w}{ v_B}[/tex]

      [tex]t_B  =   \frac{68}{ 34}  + \frac{68}{ 102}[/tex]

       [tex]t_B  =  2.67 \ hours [/tex]

The  winning boat is  boat A

The  average velocity is mathematically represented as  

      [tex]v_{avg} =  \frac{v_a  -  v_a}{ d}[/tex]

Here d is the total displacement of the winning boat which is  0 m

So

        [tex]v_{avg} =  \frac{68  -  68}{0}[/tex]

         [tex]v_{avg} =0 m/s [/tex]

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