Uniform Motion Problem: How Far is the Island?

  • Thread starter Thread starter Raizy
  • Start date Start date
  • Tags Tags
    Motion Uniform
AI Thread Summary
A ferry travels to an island at 16 mph and returns at 12 mph, with a total trip time of 2 and 1/3 hours. The distance to the island can be calculated using the formula d/v1 + d/v2 = total time. Attempts to solve the problem yielded incorrect distances, but the correct approach involves factoring out distance (d) and solving the equation d(1/16 + 1/12) = 2.33. After proper calculations, the distance to the island is confirmed to be 16 miles. Understanding the formula and proper substitution is crucial for arriving at the correct answer.
Raizy
Messages
107
Reaction score
0

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?
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top