Round-Trip Speed Word Problem: Sarah's Boat & River Current

[MarkFL]

Here is the question:

Round-Trip Speed Word Problem help?

I'm having trouble and been stuck for half an hour just trying and erasing and I need some help.

Question: In still water, Sarah's boat can travel 15 km/h. If it takes her a total of 4 1/2 hours to travel 30 km up a river and then to return by the same route, what is the speed of the current in the river?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Meh,

Let's break this down into the upstream portion of the trip and the downstream portion, using the relation distance equals speed times time. We will let $0<c$ be the speed of the current and $t_1$ be the time going upstream and $t_2$ be the time going downstream.

\(\displaystyle 30=(15-c)t_1\)

\(\displaystyle 30=(15+c)t_2\)

Now, we also know the total time is 4.5 hours, so we may state:

\(\displaystyle t_1+t_2=4.5\implies t_2=4.5-t_1\)

And so our second equation becomes:

\(\displaystyle 30=(15+c)\left(4.5-t_1 \right)\)

Solving the first equation for $t_1$, we find:

\(\displaystyle t_1=\frac{30}{15-c}\)

Now, substituting this into our new second equation, we obtain an equation in one variable $c$, which is what we are trying to find:

\(\displaystyle 30=(15+c)\left(4.5-\frac{30}{15-c} \right)\)

Multiplying through by $15-c$, we get:

\(\displaystyle 30(15-c)=(15+c)\left(4.5(15-c)-30 \right)\)

Distribute on the right:

\(\displaystyle 30(15-c)=(15+c)\left(67.5-4.5c-30 \right)\)

Combine like terms:

\(\displaystyle 30(15-c)=(15+c)\left(37.5-4.5c \right)\)

Multiply through by $2$:

\(\displaystyle 30(30-2c)=(15+c)\left(75-9c \right)\)

Distribute on both sides:

\(\displaystyle 900-60c=1125-60c-9c^2\)

Combine like terms and arrange as:

\(\displaystyle 9c^2=225\)

Divide through by $9$:

\(\displaystyle c^2=25\)

Take the positive root:

\(\displaystyle c=5\)

Thus, we conclude that the current in the river is \(\displaystyle 5\,\frac{\text{km}}{\text{hr}}\).

 

[soroban]

I'll solve this with one variable.

In still water, Sarah's boat can travel 15 km/hr.
If it takes her a total of 4 1/2 hours to travel 30 km
up a river and then to return by the same route,
what is the speed of the current in the river?

I will use: .$$\text{Time} \;=\;\frac{\text{Distance}}{\text{Speed}}$$Let $$x$$ = speed of the current.

Going upstream, her speed is $$15-x$$ km/hr.
To travel 30 km, it takes: .$$\tfrac{30}{15-x}$$ hours.

Going downstream, her speed is $$15+x$$ km/hr.
To travel 30 km, it takes: .$$\tfrac{30}{15+x}$$ hours.

Her total time is $$4\!\tfrac{1}{2}$$ hours.

. . $$\frac{30}{15-x} + \frac{30}{15+x} \:=\:\frac{9}{2}$$Multiply by $$2(15-x)(15+x)\!:$$

. . $$\begin{array}{c}60(15+x) + 60(15-x) \:=\:9(15-x)(15+x) \\ \\ 900 + 60x + 900 - 60x \:=\:9(225-x^2) \\ \\ 1800 \:=\:2025 - 9x^2 \\ \\ 9x^2 \:=\:225 \\ \\ x^2 \:=\:25 \\ \\ x \:=\:5 \end{array}$$The speed of the current is 5 km/hr.​