Tide ? "A swimmer covers a distance of 112 miles on both legs...."

[Johnny V]

A swimmer covers a distance of 112 miles on both legs . If he faces a negative tide of 3 miles per hr on the out section & the reverse on the inwards stage & takes a total of 22 hrs , what was his average speed ?

Is there a formula involving ( distance = speed * time ) that takes account of the tidal action ? 112 = ( S - 3 ) * 22 ; S = ( speed ) ? Thank you .

 

[khansaheb]

A swimmer covers a distance of 112 miles on both legs . If he faces a negative tide of 3 miles per hr on the out section & the reverse on the inwards stage & takes a total of 22 hrs , what was his average speed ?

Is there a formula involving ( distance = speed * time ) that takes account of the tidal action ? 112 = ( S - 3 ) * 22 ; S = ( speed ) ? Thank you .

the distance swam (up and down) = 112 + 112 = 224 miles

the total time swam (up and down) = 22 hrs.

Now tell us (according to your text-book) - what is the definition of "average speed" ?

 

[Steven G]

Can we see the exact problem? I believe that you might have made a mistake transcribing the problem.

 

[Dr.Peterson]

A swimmer covers a distance of 112 miles on both legs . If he faces a negative tide of 3 miles per hr on the out section & the reverse on the inwards stage & takes a total of 22 hrs , what was his average speed ?

Is there a formula involving ( distance = speed * time ) that takes account of the tidal action ? 112 = ( S - 3 ) * 22 ; S = ( speed ) ? Thank you .

Presumably what they are asking for is not his average actual speed, but his average speed relative to the water. That is, if his speed in still water is always x miles per hour, then swimming against the tide his actual speed is x-3, and swimming with the tide it is x+3.

Also, presumably the distance is 112 for each leg, not for both legs combined.

I get a reasonable answer interpreting the problem this way, though the wording could be a lot clearer.

 

[limiTS]

If asking for his speed [imath]S[/imath] relative to the water, then set it up like this:

[imath]\displaystyle\frac{112\text{ mi}}{(S-3)\text{ mi/h}}+\frac{112\text{ mi}}{(S+3)\text{ mi/h}}=22\text{ h}[/imath]

Solve for [imath]S[/imath]