Uniform Motion Problem: How Far is the Island?

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Homework Help Overview

The problem involves a ferry traveling to an island at different speeds for the outbound and return trips, with a total travel time provided. Participants are tasked with determining the distance to the island based on the given speeds and total time.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various attempts to set up equations based on the relationship between distance, speed, and time. Some suggest expressing the total time as the sum of individual travel times, while others explore different algebraic manipulations to isolate the distance variable.

Discussion Status

There are multiple approaches being explored, with some participants questioning the accuracy of their calculations and others providing alternative methods to express the problem. Guidance has been offered on how to set up the equations, but no consensus on a final solution has been reached.

Contextual Notes

Participants note the total travel time of 2 and 1/3 hours and the different speeds for each leg of the trip. There is an acknowledgment of potential errors in calculations and the importance of verifying equations before concluding.

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Homework Statement



A ferry leaves a dock and travels to an island at an average speed of 16 mph. On the return trip, the ferry travels at an average speed of 12 mph. The total time for the trip is 2 and 1/3 hours.

How far is the island from the dock?

The Attempt at a Solution



Attempt 1:

d = distance

(7/3 * 16) - (7/3 * 12 ) = d
112/3 - 84/3 = d
28/3 = d
d = 9.333...

Attempt 2:

t = time of second trip

16 * (2 and 1/3 - t)= 12t
37 and 1/3 - 16t = 12t
37 and 1/3 = (12 + 16)t
112/3 = 38t
1/38 * 38/1t = 1/38 * 112/3
t = 112/114
t = 56/57

If I use 56/57 I get 21.61 mi and 11.79 mi. lol.

The book's answer is 16 miles.

Any takers?
 
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Well since you know the total time for the trip, try starting with this idea

t1 + t2 = ttotal

You can easily express t1 and t2 in other variables.
 
It is motion at constant speed, so you use d = vt for each part of the trip.
You know that the time for the first part plus the time for the second part is 2.33 hr, so it makes sense to solve the formula for t = d/v.

1st time + second time = 2.33 hr
d/v1 + d/v2 = 2.33
You'll have to factor the d out to solve for it.
 
Holy crap my clumsy ***'s attempt 2 approach was right all along.

Damn, I must be so wrong always forgetting to double check my equations.

t = time of second trip

16 * (2 and 1/3 - t)= 12t
37 and 1/3 - 16t = 12t
37 and 1/3 = (12 + 16)t
112/3 = 38t It's supposed to be 28t ! *sigh*
1/38 * 38/1t = 1/38 * 112/3
t = 112/114
t = 56/57

After plugging in t = 1.33 to calculate distance it's pretty close to 16 mi.
 
I can't follow your work! Aren't you supposed to write a formula, then sub the numbers in and solve for the unknown? It turns out that answers alone are worth nothing in the marketplace - rather, you have to present a solution that convinces your employers that you know what you are talking about before they will take note of your answer!

1st time + second time = 2.33 hr
d/v1 + d/v2 = 2.33
You'll have to factor the d out to solve for it.
d/16 + d/12 = 2.33
d(1/16 + 1/12) = 2.33
Run that bracket through your calculator to get a number for it.
Divide both sides by that number to get a d = answer.
 

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