Ellie S.
asked 07/21/17murphy can row at 5kmph in still water. If the velocity of the current is 1kmph and it takes him 1 hour to row to the place and back, how far is the place?
A question from physics homework that is really confusing me
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2 Answers By Expert Tutors

Walter B. answered 07/21/17
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Let's breakdown the problem into an algebra problem.
the distance from the place when Murphy rows with the current is 6kmph (5 + 1), and is 4kmph (5 -1) when Murphy rows against the current.
Let's define two new variables t1 and t2 as the times to row to the place and back, respectively.
the distance, in each case, = velocity * time = 6* t1 = 4 * t2 because distance is the same. Also we know that t1 +t2 = 1hour
Now we have our two equations
t1 + t2 = 1
6t1 = 4t2 or t1 = (2/3)t2
2/3t2 + t2 = 1
5/3t2 = 1
t2 = 3/5
d = t2 * v2 = t1 * v1 = (3/5) * 4 = (2/5) * 6
Can you finish the calculation?

Martin P. answered 07/21/17
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From Educator to Engineer
Murphy rows 5kmph in still water. (Wow, 5000 mph.. now that is fast.. He could circle the Earth in 5 hours at that speed... whew... ) Since these speeds are not reasonable, I will change them to more reasonable speeds by dropping the k off each. The current pushes back his forward progress at a rate of 1mph. In effect he paddles upstream at a rate of 4mph due to paddling against the current. Paddling with the current (coming back) he paddles 6mph.
Forming two equations with two unknown. We know the time spent paddling upstream is not the same as the time spent paddling downstream because if it were, he would not end up at the same spot he started.
But we do know the total time he spends traveling upstream at 4kmph and the total time he spends traveling downstream at 6mph is 1 hour. Our equation looks like.
t4 + t6 = 1 (hour)
t4 = 1 - t6 [Solving or t4, the time he spends paddling upstream.] Hint: He should spend more time paddling up stream than down stream since both are the same distance and he travel rate is slower upstream.
We also know that the distance he paddles upstream is the same distance he paddles downstream, giving us the equation.
4[mi/hr] * t4[hr] + 6[mi/hr] *t6[hr] = total distance
4[mi/hr] * t4[hr] = 6[mi/hr] *t6[hr] [Setting both distances equal to each other.]
4[mi/hr] * (1-t6)*t4[hr] = 6[mi/hr] *t6[hr] [Plugging in t4 equivalent of t4 = 1-t6.]
t6 = 4/10
t4 = 1 - t6 = 6/10 [Solving for t4 given t6. Notice he spends more time paddling upriver.]
Plugging t4 and t6 into the total distance equation we get
4[mi/hr] * 6/10 [hr] + 6[mi/hr] *4/10[hr] = 48/10 = 4.8 miles total distance
The spot where he traveled to is 2.4 miles.
*Remember I removed the "k" from the distance because that was just ridiculous in this application.
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