Help with homework?

  • Thread starter phywhat
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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?
 

Answers and Replies

  • #2
this should go in the homework section... as well... you should show some work first, otherwise ppl are not going to help you.
 
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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 [itex]v_A = 3[/itex] and [itex]v_B = 4[/itex] Integrating with respect to time gives [itex]x_A = 3t + C_A[/itex] and [itex]x_B = 4t + C_B[/itex]. When [itex]t=0[/itex], [itex]x_A = 0[/itex] so [itex]x_A = 3t[/itex]. When [itex]t=300[/itex], [itex]x_B = 0[/itex], so [itex]x_B = 4t - 1200[/itex]. Runner B catches runner A when [itex]x_A=x_B[/itex]. That should be all you need to know.
 

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