Question

a boat travels 33 miles Downstream in 4 hours. The return trip takes the boat 4 hours. Find the speed of the boat in still water.

Answer 1

Let's use the following formula:

$v=\frac{d}{t}$

We have two speeds in this case, the speed of the boat and the speed of the water, so:

$\begin{gathered} v_b=Speed_{\text{ }}of_{\text{ }}the_{\text{ }}bo_{}at \\ v_w=Speed_{\text{ }}of_{\text{ }}the_{\text{ }}water \end{gathered}$

When the boat travels 33 miles downstream in 4 hours, we can say that their speeds add up:

$v_b+v_w=\frac{33}{4}$

When the boat return, we can say that their speeds are subtracted:

$v_b-v_w=\frac{33}{4}$

with this we can form a system of equations 2x2:

$\begin{gathered} v_b+v_w=\frac{33}{4}_{\text{ }}(1)_{} \\ v_b-v_w=\frac{33}{4}_{\text{ }}(2) \end{gathered}$

Let's solve for vb:

$\begin{gathered} (1)+(2) \\ v_b+v_b+v_w-v_w=\frac{33}{4}+\frac{33}{4} \\ 2v_b=\frac{66}{4} \\ 2v_b=\frac{33}{2} \\ v_b=\frac{33}{4}=\frac{8.25mi}{h} \end{gathered}$