Solve System of Eqs: Miles, Time, Boat, Current

[NoPhysicsGenius]

[SOLVED] System of equations

1. Homework Statement

Use a system of equations to solve the following problems:

(a.) A boat took 1 hour 50 minutes to go 55 miles downstream and 3 hours 40 minutes to return. Find the rate of the current.

(b.) A boat travels 60 miles downstream in the same time it takes to go 36 miles upstream. The speed of the boat is 15 mi/h greater than the speed of the current. Find the speed of the current.

(c.) A boat travels 12 miles downstream in 1.5 hours. On the return trip the boat travels the same distance upstream in 2 hours. Find the rate of the boat in still water and the rate of the current.

2. Homework Equations

distance = rate * time

3. The Attempt at a Solution

Here's a lame-brained attempt to solve problem (b) ...

[tex]d=rt[/tex]
[tex]r=r_b + r_w[/tex]
[tex]d = (r_b + r_w)t[/tex]
[tex]r_b = 15 + r_w[/tex]
[tex]d = (15 + 2r_w)t[/tex]
? ...

I can't figure out how to solve any of these. Could you at least help me to set them up correctly? Thank you.

 

[symbolipoint]

Note in part b that the two directions of trips took equal times, t.

Some of your starter relations and expressions are:

[tex] \[
\begin{array}{l}
r_{b\;} \,speedOfBoat \\
r_w \;speedOfWater \\
alsoGivenIsThat\;\,r_b = 15 + r_w \\
\end{array}
\]
[/tex]

 

[symbolipoint]

...woops, I did not yet finish:

[tex] \[
\begin{array}{l}
downStreamRate,\quad r_b + r_w \\
upStreamRate,\;\;\quad r_b - r_w \\
\end{array}
\]
[/tex]

...just to help you begin.

 

[HallsofIvy]

1. Homework Statement

Use a system of equations to solve the following problems:

(a.) A boat took 1 hour 50 minutes to go 55 miles downstream and 3 hours 40 minutes to return. Find the rate of the current.

(b.) A boat travels 60 miles downstream in the same time it takes to go 36 miles upstream. The speed of the boat is 15 mi/h greater than the speed of the current. Find the speed of the current.

(c.) A boat travels 12 miles downstream in 1.5 hours. On the return trip the boat travels the same distance upstream in 2 hours. Find the rate of the boat in still water and the rate of the current.

2. Homework Equations

distance = rate * time

3. The Attempt at a Solution

Here's a lame-brained attempt to solve problem (b) ...

[tex]d=rt[/tex]
[tex]r=r_b + r_w[/tex]
[tex]d = (r_b + r_w)t[/tex]
[tex]r_b = 15 + r_w[/tex]
[tex]d = (15 + 2r_w)t[/tex]
? ...

Unfortunately you don't say what "r_b" and "r_w" mean, so it is impossible to say whether those are correct or not! (Yes, I can guess that "r_b mean "rate of boat" and "r_w" means "rate of water" but it would have been better to say that! Especially since in the problems you used "rate" and "speed" interchangeably and you really mean "speed of the current", not "rate of the water"! Also, you don't say anything about how you got those equations.

I can't figure out how to solve any of these. Could you at least help me to set them up correctly? Thank you.

(a.) A boat took 1 hour 50 minutes to go 55 miles downstream and 3 hours 40 minutes to return. Find the speed of the of the current.

Let v_b be the speed of the boat in still water and v_c the speed of the current, both in miles per hour. Going upstream the speed of the boat relative to the bank is v_b- v_c. In 3 hours and 40 minutes = 3 2/3= 11/3 hours, the boat will go (11/3)(v_b-v_c)= 55 miles upstream. Going down stream the boats speed, relative to the bank is v_b+ v_c. In an hour and 50 minutes= 1 5/6= 11/6 hours the boat will go (11/6)(v_b+ v_c)= 55. Those are the two equations you need to solve for v_c.

(b.) A boat travels 60 miles downstream in the same time it takes to go 36 miles upstream. The speed of the boat is 15 mi/h greater than the speed of the current. Find the speed of the current.

Again, let v_b be the speed of the boat in still water, and let v_c be the speed of the current, both in miles per hour. Again, the speed of the boat relative to the bank, going upstream is v_b- v_c and the speed of the boat goiong downstream is v_b+ v_c. Since "the spead of the boat is 15 mi/h greater than the speed of the current, v_b= v_c+ 15. vb-v_c= vc+ 15- vcc= 15 mi/h. v_b+ vc= 15+ v_c+ vc= 15+ 2vc. It would take 60/(v_b- v_c)= 36/15= 12/5 hours to go 36 miles upstream. It would take 60/(vb+ v_c)= 60/(15+ 2v_c hours to go 60 miles downstream. Since those times are the same, 12/5= 60/(15+ 2v_c). That is the equation you need to solve for v_c.

(c.) A boat travels 12 miles downstream in 1.5 hours. On the return trip the boat travels the same distance upstream in 2 hours. Find the rate of the boat in still water and the rate of the current.

Again, let v_b be the speed of the boat in still water, and let v_c be the speed of the current, both in miles per hour. Again, the speed of the boat relative to the bank, going upstream is v_b- v_c and the speed of the boat goiong downstream is v_b+ v_c. If the boat travels 12 miles downstream in 1.6 hours, then its speed, relative to the bank, is 12/1.5= 8 mi/h= v_b+ v_c. Traveling the same distance, 12 miles, in 2 hours, its speed, relative to the bank, is 12/2= 6 mi/h= v_b- v_c. Those are the two equations you can solve for v_b and v_c.

 

[symbolipoint]

Part b is best handled with a data table: two rows marked as Up Stream and Down Stream; start with three columns marked as Rate(or Speed), Time, Distance. Create more columns as you find useful. You will eventually equate the times for up and down stream.

 

[NoPhysicsGenius]

Thanks for the help, guys! (Especially HallsofIvy!)

 

[rocomath]

Thanks for the help, guys! (Especially HallsofIvy!)

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