Speed of Boat & Current Rate

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mathdad

mathdad

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Traveling downstream, Elmo can go 6 km in 45 minutes. On the return trip, it takes him 1.5 hours. What is the boat's speed in still water and what is the rate of the current?

Let b = boat
Let c = current

Downstream: b + c
Upstream: b - c

Note: 45 minutes = 3/4 of an hour or 0.75 hour.

Total distance = 6 km

Equation 1

(3/4)(b - c) = 6

Equation 2

1.5(b - c)

Is the set up correct?
 
No, but given your correct setups on your several other nearly identical problems, I think you should simply examine what you've posted here and try again. :)
 
MarkFL said:
No, but given your correct setups on your several other nearly identical problems, I think you should simply examine what you've posted here and try again. :)
Click to expand...

Let b = boat
Let c = current

Downstream: b + c
Upstream: b - c

Note: 45 minutes = 3/4 of an hour or 0.75 hour.

Total distance = 6 km

Equation 1

(3/4)(b + c) = 6

Equation 2

1.5(b - c) = 6

How about now?
 
mathdad said:
Let b = boat
Let c = current

Downstream: b + c
Upstream: b - c

Note: 45 minutes = 3/4 of an hour or 0.75 hour.

Total distance = 6 km

Equation 1

(3/4)(b + c) = 6

Equation 2

1.5(b - c) = 6

How about now?
Click to expand...

Yes, that looks good now. 8-)
 
MarkFL said:
No, but given your correct setups on your several other nearly identical problems, I think you should simply examine what you've posted here and try again. :)
Click to expand...

What about this one?

Into the headwind, the plane flew 2000 miles in 5 hours. With a tailwind, the return trip took 4 hours. Find the speed of the plane in still air and the speed of the wind.

Solution:

Let p = speed of plane
Let w = speed of wind

I need to use d = rt.

Into headwind rate: p - w
With the wind rate: p + w

Going Trip Time: 5 hours
Returning Trip Time: 4 hours

Total distance: 2000

Set up:

5(p - w) = 2000
4(p + w) = 2000

Is this correct?

I say yes.
 
You should note that all of the problems of this type, where we have some vehicle traveling through/on a fluid medium with/against a current, that we will obtain a system line:

[MATH]d=(v+c)t_1[/MATH]
[MATH]d=(v-c)t_2[/MATH]
What I did as a student was solve that system for \(v\) and \(c\) in terms of the other parameters, to obtain formulas to apply to all such problems. :)
 
MarkFL said:
You should note that all of the problems of this type, where we have some vehicle traveling through/on a fluid medium with/against a current, that we will obtain a system line:

[MATH]d=(v+c)t_1[/MATH]
[MATH]d=(v-c)t_2[/MATH]
What I did as a student was solve that system for \(v\) and \(c\) in terms of the other parameters, to obtain formulas to apply to all such problems. :)
Click to expand...

Ok but is my set up right?
 
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