Swimmer pool meeting

The pool is 50 m long. The first swimmer swims across it in the 60 s and the second in the 40 s. Both start at the same time and on the same side. The faster swimmer turns around at the end and swims back. When will he meet the slower swimmer? Swimmers move at constant speeds and do not consider the turn time.

Final Answer:

t =  48 s

Step-by-step explanation:

x=50 m t1=60 s t2=40 s  v1=x/t1=50/60=650.8333 m/s v2=x/t2=50/40=45=1.25 m/s  s1+s2=x+x  t v1+t v2=2 x t 0.83333333333333+t 1.25=2 50  2.083333t=100  t=2.08333333100=48  t=48=48 s



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