Consider 82380

The pool is 50m long. The first swimmer swims across it in the 60s and the second in the 40s. 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.

Correct 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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