Emily rows six miles downstream in 1 hour and her friend Ashley, rowing 1 mile per hour faster, completes the return trip in 2 hours.
Find the speed of the current (c) and each girls rowing speed.
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We assume here that the speed of the current is simply added or subtracted to the speed of interest if the object is flowing along or against the current, respectively.
We only need to know that `d = vt` , where v is the speed, and t is the time.
We let X to be the speed of Emily (this makes Ashley's X+1).
Then, we let V to be the speed of the current.
Distance travelled by Emily (travelled in an hour) is given by:
`d = (X + V)*1 = X + V`
Distanced travelled by Ashley (in two hours) is:
`d = (X + 1 - V)*2 = 2X + 2 - 2V`
Note that Emily is travelling downstream so we added the speed of the current, while Ashley is travelling upstream (against the current) so we subtract V.
We know that both of them travelled 6 miles.
Hence we have a system of two equations, namely:
`X + V = 6`
`2X - 2V + 2 = 6 \rightarrow 2X - 2V = 4 \rightarrow X - V = 2`
Adding both equations gives us:
`X + X + V - V = 8`
`2X = 8 \rightarrow X = 4`
Substituting back to the first equation:
`4 + V = 6 \rightarrow V = 6-4 = 2`
Emily's speed is 4 miles/hour downstream, Ashley's is (4+1) = 5miles/hour upstream, and the current's speed is 2 miles/hour.
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