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Question

1) len is planning a three hour trip own the Allemen River and back to his starting point. He knows that he can paddle in still water at 3 miles per hour and that the rate of the current is 2 mph. how much time can he spend going downstream? how far downstream can he travel?

2)a motorboat has a four hour supply of gasoline. how far from the marina can it travel if the rate going out against the current is 20 mph and the rate coming back with the current is 30 mph?

3)an airplane whose speed in still air is 760 kmp can travel 2000 km with the wind in the same amount of time that it takes to fly 1800 km against the wind. What is the speed of the wind?

i really don't get how to do these. can someone solve and explain please?

Answer 1

Let the speed of the winfd be x.
Speed of plane with the wind=760+x
Speed of plane aganst the wind=760-x

Speed=Distance/Time
Therefore
Time=D/S
2000/760+x=y
1800/760-x=y


y=time taken.Plus its given that the same amount of time in both the cases.U can solve these 2 equations.and u will ge the answer.

Answer 2

1. Use the distance formula, D = RT, and assign paddle speeds up and down the river and times traveled in each direction. On the downstream leg, since the current is going in the same direction, its speed is added to his paddling speed. On the way back, the current is opposing his paddling, so it is subtracted from his speed. The distances each way are equal.

T(1) = time traveling downstream
T(2) = time traveling upstream

T(1) + T(2) = T (total time) = 3 hours

R(1) = speed paddling downstream = (3 + 2) mph
R(2) = speed paddling upstream = (3 - 2) mph

D(1) = distance traveled downstream = distance traveled downstream = D(2).

Because D(1) = D(2), then R(1)T(1) = R(2)T(2).

Since T(1) + T(2) = T = 3, then we can solve for the time it takes to return, T(2), in terms of the time he spends paddling downstream, T(1):

T(2) = T - T(1) = 3 - T(1).

Now we can plug T(1) and our calculated value for T(2) into our distance formula and solve for T(1).

R(1)T(1) = R(2)T(2)
[(3 + 2) mph] T(1) = [(3 - 2) mph] [3 - T(1)]
5 mph T(1) = 1 mph [3 - T(1)]
[(5 + 1) mph] T(1) = 3 mph
6 mph T(1) = 3mph
T(1) = (3/6) hours = 1/2 hour

So T(2) = 3 hours - T(1) hours

T(2) = 3 hours - (1/2) hour = 5/2 hours.

To check the answers, substitute these times into the distance formula and see whether D(1) = D(2). If they do, then we have the right numbers.

(1/2) hour x 5 mph = (5/2) hours x 1 mph.

(5/2) miles = (5/2) miles: These numbers check out.

So Len can paddle 5/2 or 2.5 miles downstream before he must turn around and go back.

#2. Remember, the distances going each way must be equal, so set D(1) = D(2), and here T(1) + T(2) = 4 hours. By solving for T(2) in terms of T(1) like we did in the first problem, you can find the actual value of T(1) by plugging both into the distance formula:

R(1)T(1) = R(2)T(2) ----> R(1)T(1) = R(2)[T - T(1)]
R(1)T(1) = R(2)T - R(2)T(1)
R(1)T(1) + R(2)T(1) = R(2)T
T(1)[R(1) + R(2)] = R(2)T
T(1) = R(2)T / [R(1) + R(2)]

Then simply multiply R(1), 20 mph, by T(1) to find D(1). Make sure that you also calculate the actual value of T(2) to verify that D(2) equals D(1). If it doesn't, something is amiss

#3. Here T(1) = T(2) = 760 kph / 2000 km = 38/100 hours. You are given the speed in still air. But the distance traveled in still air is 200 km more than when flying against the wind, so D(1) = D(2) + 200 km

R(1)T(1) = R(2)T(2) + 200 km and R(1) = 760 kmp. So

760 kph T(1) = R(2)T(2) + 200 km

At this point, you can find R(2) easily since T(1) = T(2). Then just compare that to the speed of the airplane in still air. That will tell you the speed of the wind:

R(2) = R(1) - Wind speed
R(2) = 760 kmp - Wind speed
760 mph - R(2) = Wind speed.

In the last line of the above calculation, we subtract R(2) from 760 because it is less than the speed in still air. Had we subtracted the terms in the reverse order, it would have told us that the wind was having a negative effect on the airspeed, since it would be a negative number. But we want only the actual speed of the wind, not its net effect.

.

Answer 3

In word problems you have to read the question and find out what they want to know. Then you have to look at the information that is given and see how you can use it to find the answer.

Question 1

You want to know how far the guy is going to be able to go down the river and make it back in three hours.

He starts out going down the river then turns around and comes back up.

There is a current in the river so he is going to have to spend more time paddling upriver against the current than he will going down the river with the current.

His total time has to be three hours.

He paddles at three miles an hour where there is no current (still water).

We will assume that he is on the river he is always paddling at three miles per hour, whether he is going up or down.

We need to realize logically that he will be going down the river the same distance as he will be going up but that going back up will take longer.

So if the river is flowing at 2 miles per hour and he is paddling at three miles per hour then his total speed down the river is 2+3=5 miles per hour.

Going back up the river he is going against the current so his speed is 3 miles per hour minus 2 miles per hour from the current or 3 - 2 = 1 mile per hour.

So how far can he go down the river and still be able to get back in time.

We realize that he travels down the river 5 times as fast as he goes up 5 mph vs 1 mph.

So he's going to have to spend 5 times as long going back as he did going down.

lets say that x is the amount of time he spends going down, then 5x is the amount of time he spends going back.

So the total amount of time is 5x + 1x = 6x

We know that the total time is 3 hours. So 6x = 3 hours.

if we solve for x we have 6x = 3 and divide both sides by 6 we have x = 3/6 and 3/6 is the same as 1/2.

so x = 1/2

so he spends 1/2 hour going down the river and 2 and 1/2 going back up.

at 5 miles an hour 1/2 hour travel is 2.5 miles.

So our answer to the problem is that he spends one half hour going down river and travels 2 and one half miles in the process.

I tried to break this problem down into parts that are easier to understand. See if you can apply this method to the other problems. The more practice you have the better you will be.

Answer 4

eww. we did these in algebra last year. just remember distance equals rate times time. d=rt
try to think of it logically.. try really hard. :)

good luck