Runner A & B Race: Find Out When & How Far!

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Runner A starts running at 3:00 P.M. with a speed of 3.0 m/s, while Runner B begins 5 minutes later at a speed of 4.0 m/s. To determine when Runner B catches Runner A, the equations for their distances can be set equal: x_A = 3t and x_B = 4(t - 300). Solving for t reveals that Runner B catches Runner A at a specific time, which can be calculated from their speeds and starting times. The distance they run until B catches A can also be calculated using these equations. This problem illustrates the application of relative speed and time in motion scenarios.
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Help with homework??

Runner A, who runs with an average speed of 3.0 m/s, starts out at 3:00 P.M. Runner B, who runs with an average speed of 4.0 m/s, starts after A from the same place exactly 5 min later.
a.) At what time will runner B catch up with runner A?
b.) If the runners stop when B catches A, how far do they run?
 
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this should go in the homework section... as well... you should show some work first, otherwise ppl are not going to help you.
 
phywhat said:
Runner A, who runs with an average speed of 3.0 m/s, starts out at 3:00 P.M. Runner B, who runs with an average speed of 4.0 m/s, starts after A from the same place exactly 5 min later.
a.) At what time will runner B catch up with runner A?
b.) If the runners stop when B catches A, how far do they run?
You know v_A = 3 and v_B = 4 Integrating with respect to time gives x_A = 3t + C_A and x_B = 4t + C_B. When t=0, x_A = 0 so x_A = 3t. When t=300, x_B = 0, so x_B = 4t - 1200. Runner B catches runner A when x_A=x_B. That should be all you need to know.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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