--> total distance traveled
so now we divide by 2 to get 6, how far he traveled in the forest.
The question is:
The answer is 6??A man walked into the woods at the rate of 4 miles per hour and returned over the same rate at 3 miles per hour.If he completed the entire trip in 3.5 hours how far in miles did he walk into the woods ?
Any idea how i would solve this ???
t= (3/7)(3.5) = 1.5 this is the time it with have taken travelling at 4 miles per hour, we take (3/7) of total time because that is the percentage of hours it takes to travel half the distance, this number will be less then the return journey, since our return journey speed was less.
speed = 4 miles per hour
distance traveled = speed * time = 1.5 * 4 = 6
Goku explain please strictly mathematical.
First you need to know that man traveled a fix distance to the wood, d. He traveled this distance d with different speed when he go and when he returned.
So:
d = V1 * t1
d = V2 * t2
t1+t2 = 3.5 hours. We need to solve d, but in order to do that, we need to find out t1 or t2.
V1*t1=V2*t2 => 4*t1=3*t2 => t1=(3/4)*t2
t1+t2=3.5 <=> (3/4)*t2 + t2 = 3.5 => t2 = 3.5 * (4/7) = 2 hours
So it results from first equation:
d = V2 * t2 = 3 miles/hour * 2 hours = 6 miles
let t = time walking into the woodsA man walked into the woods at the rate of 4 miles per hour and returned over the same rate at 3 miles per hour.If he completed the entire trip in 3.5 hours how far in miles did he walk into the woods ?
3.5 - t = time walking out of the woods
distance walked in = distance walked out
4t = 3(3.5 - t)
solve for t, then find the distance into the woods ... 4t.
Let = t1 wood time and = t2 returning time
Now we're going to use the distance-rate-time equation d = rt
Since the walking speed is 4 mph and the wood time is t1, this means r = 4 and t = t1
d = 4t1 Plug in r =4 and t = t1
Also, since the returning speed is 3 mph and the driving time is t2, this means r = 3 and t = t2
d = 3t2 Plug in r= 3 and t= t2
Because the distance is the same both ways, this means that d = 4t1 = 3t2
4t1 = 3t2 Start with the given equation
t1 = 3/4 t2 Divide both sides by 4 to isolate t1
entire trip took 3.5 hour
So the sum of these two times is 3.5. In other words, we have this first equation
t1 + t2 = 3.5
3/4 t2 + t2 = 3.5 plug in t1 = 3/4 t2
1.75 t2 = 3.5 Combine like terms
t2 = 2 Divide both sides by 1.75 to isolate t2
so walking time is 2 hour
Now let's go back to the equation d= 3t2
d = 3t2 Start with the given equation
d = 3 * 2
d = 6 miles
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skeeter's solution is probably the best so far. If the man takes hours to walk into the woods, he will take hours walking out of the woods. We know that the distances into and out of the woods are obviously equal, so
(hours)
Hence, so the man walked 6 miles into the woods. As a check, as well.
Hello, MikeNoob!
A man walked into the woods at the rate of 4 mph and returned over the same path at 3 mph.
If he completed the entire trip in 3.5 hours, how far in miles did he walk into the woods?
The answer is 6
Any idea how i would solve this ?
Since it mentions Distance, Speed and Time, you might suspect that
. . we could use a formula involving these three quantities.
How about: .
The man walked miles into the wood at mph.
. . This took: hours.
Then he walked miles out of the woods at mph.
. . This took: hours.
His total time is: hours, which is hours.
There is our equation! . . . .
Now solve for