Question

Running at maximum speed, it takes a boat times as long to go 10 miles upstream as it does to go 10 miles downstream. If the current is 4 mph, find the speed of the boat in still water.

Answer 1

Note: The question states the time to go upstream is a number of times (not explicitly written) the time to go downstream. We'll assume a general number N

Answer:

$\displaystyle v_b=\frac{N+1}{N-1}(4\ mph)$

Explanation:

Relative Speed

If a boat is going upstream against the water current, the true speed of motion is $v_b-v_w$, being $v_b$ the speed of the boat and $v_w$ the speed of the water. If the boat is going downstream, the true speed becomes $v_b+v_w$.

The question states the time to go upstream is a number of times N (not explicitly written) the time to go downstream. The speed of an object is computed as

$\displaystyle v=\frac{x}{t}$

Where x is the distance traveled and t the time taken for that. The time can be computed by

$\displaystyle t=\frac{x}{v}$

If $t_u$ is the time for the upstream travel and $t_d$ is the time for the downstream travel, then

$t_u=Nt_d$

Siince the same distance x= 10 miles is traveled in both cases:

$\displaystyle \frac{10}{v_b-v_w}=N\frac{10}{v_b+v_w}$

Simplifying and rearrangling

$v_b+v_w=N(v_b-v_w)$

Operating

$v_b+v_w=Nv_b-Nv_w$

Solving for $v_b$

$\displaystyle v_b=\frac{N+1}{N-1}v_w$

$If\ N=2,\ v_w=4\ mph$

$\displaystyle v_b=\frac{3}{1}(4)=12\ mph$

If N=3

$\displaystyle v_b=\frac{4}{2}v_w=2(4)=8\ mph$

We can use the required value of N to compute the speed of the boat as explained