How do you do a Round-Trip Distance Word Problem?

Question

I have just begun classes this fall and I seem to be a bit behind other students in my Algebra 2 class. I'm lost...

These are both sample problems from our text books, so don't worry, I won't receive a grade for YOUR work.

The problems read :

1. A ferry travels from the dock to a cruise ship at a speed of 40 miles per hour and returns at a speed of 25 miles per hour. If the entire trip took 13 hours, how long did it take the ferry to travel from the dock to the cruise ship?

(The one above I do not know how to show my work on)

2. Jim can fly his airplane 3.825 kilometers with the wind in the same time it takes him to fly 3675 kilometers against the wind. If the speed of the wind is 10 kilometers per hour, find the speed of the airplane in still air.

The main problems I'm having is working the problems out...help would be greatly appreciated.

Answer 1

These are Rate x Time = Distance problems.

1. Here we do not know the distance,

but we do have information about Rate and Time.

Outbound time = t1, Return time = t2.  

t1 + t2 = 13  (total time is 13 hours)

40 x t1 = 25 x t2 (distance is the same each way)

40 (13 - t2) = 25 x t2

520 - 40 t2 = 25 t2

520 = 65 t2

t2 = 520 / 65 = 8 hours

t1 = 5 hours

Distance is 200 miles.

2.

Now we have distance and time information, but not speed.

Let s = speed of plane

With the wind he goes s + 10

against the wind he goes s - 10

The time is the same, call it t

t (s + 10) = 3825

t (s - 10) = 3675

st + 10t = 3825

st - 10t = 3675

Subtract to get rid of the "st" terms

20 t = 150

t = 7.5 hours

7.5 (s + 10) = 3825

s+10 = 3825/7.5 = 510

s - 10 = 3675/7.5 = 490 (just to check)

So s = 500.

I find that hitting the ENTER key once in a while improves readability.

Do you agree ?

.

Answer 2

1.  What you need to do here is call the amount of time it takes on the first leg X, then you know the second leg takes 13 - X.  Since speed times time equals distance and you know both trips are the same distance, you can say 40 mph * X = 25 mph * (13 - X) or 40X = 25(13 - X) or 40X = 325 - 25X.  Now add 25X to each side to get 65X = 325 or 5 hours.  It took the ferry 5 hours to get out and 13 - X or 13 - 5 or 7 hours on the return trip.

2. First I'm assuming 3.825 km is actually supposed to be 3825 km.  Call Jim's speed in still air Y, so when flying with the wind he is flying at Y + 10 and against the wind he is flying at Y - 10.  Again, speed times time = distance.  Dividing both sides by speed, you get time = distance / speed.  Since the time is the same in both cases, you can say 3825/(Y + 10) = 3675/(Y-10).  Multiply both sides by (Y + 10) then multiply both sides by (Y - 10) to get 3825(Y - 10) = 3675(Y + 10) or 3825Y - 38250 = 3675Y + 36750.  Subtract 3675Y from both sides, then add 38250 to both sides to get 150Y = 75000 or Y = 500 km per hour

Answer 3

GregH13

1. A ferry travels from the dock to a cruise ship at a speed of 40 miles per hour and returns at a speed of 25 miles per hour. If the entire trip took 13 hours, how long did it take the ferry to travel from the dock to the cruise ship?

Let D = distance from ferry to cruise

Speed ferry travels from the dock to a cruise ship = 40 mph

Speed ferry travels from the dock to a cruise ship = 25 mph

Total time for the entire trip = 13 hours

Speed = Distance / Time  ÃƒÂ¢Ã‚†Â Formula

Time = Distance / Speed

Time to Cruise Ship + Time from Cruise = Total Time

D/40 + D/25 = 13

25D + 40D = (25)(40)(13)

65 D = 13000

D =  200 miles

2. Jim can fly his airplane 3825 kilometers with the wind in the same time it takes him to fly 3675  kilometers against the wind. If the speed of the wind is 10 kilometers per hour, find the speed of the airplane in still air.

Let S = Speed for the Plane

T = time spent  for the trip

Speed of the Wind = 10 km/hr

Speed with the Wind = (S + 10) → km/hr

Speed against the Wind = (S – 10)→ km/hr

Distance = Speed x Time

Distance With the Wind = 3825 km     3675 km

Distance against the Wind =  3675 km

So (S + 10) ← km/hr x Time (T) => distance with the Wind ← eq 1

And  (S - 10) ← km/hr x Time (T) => distance against the Wind ← eq 2

T(S + 10) = 3825 ← eq 1

TS + 10 T = 3825 ← eq 3

T(S - 10) = 3675 ← eq 2

TS - 10T = 3675 ← eq 4

Multiply eq 4 by (-1)

-TS + 10T = -3675 ← eq  5

Add algebraically eq 3 & eq 5

Result will be

20T =  150

T = 7.5 hour

Subsititutte T = 7.5 to eq 1

T(S + 10) = 3825 ← eq 1

7.5(S + 10) = 3825

7.5S + 75 = 3825

7.5S = 3750

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Ans:: S = Speed of the Plane = 500  km/hr

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Hope this helps