# Math Help - Motorboat

1. ## Motorboat

A motorboat can go 16 miles downstream in 20 minutes. It takes 30 minutes for this same baot to go upstream the same 16 miles.

Let x = the speed of the boat.
Let y= the speed of the current.

a. Write an equation for the motion of the motorboat downstream.
b. Write an equation for the motion of the motorboat upstream.
c. Find the speed of the current.

2. Let d = 16 miles
Let $t_d = 20$ minutes
Let $t_u = 30$ minutes

The speed of the boat when it goes downstream is x+y
Therefore $\frac{d}{x+y} = t_d$
Inverse gives (i) $\frac{x}{d} + \frac{y}{d} = \frac{1}{t_d}$

The speed of the boat when it goes upstream is x-y
Therefore $\frac{d}{x-y} = t_u$
Inverse gives (ii) $\frac{x}{d} - \frac{y}{d} = \frac{1}{t_u}$

Equation (i) - equation (ii) gives
$2 \frac{y}{d} = \frac{1}{t_d} - \frac{1}{t_u}$

$y = \frac{d}{2} (\frac{1}{t_d} - \frac{1}{t_u})$

y = 8 mph

3. $v=\frac{s}{t}$

$x+y=\frac{16}{20}$

$x-y=\frac{16}{30}$

Then you have two simultaneous equations to solve.

4. ## ok....

Originally Posted by running-gag
Let d = 16 miles
Let $t_d = 20$ minutes
Let $t_u = 30$ minutes

The speed of the boat when it goes downstream is x+y
Therefore $\frac{d}{x+y} = t_d$
Inverse gives (i) $\frac{x}{d} + \frac{y}{d} = \frac{1}{t_d}$

The speed of the boat when it goes upstream is x-y
Therefore $\frac{d}{x-y} = t_u$
Inverse gives (ii) $\frac{x}{d} - \frac{y}{d} = \frac{1}{t_u}$

Equation (i) - equation (ii) gives
$2 \frac{y}{d} = \frac{1}{t_d} - \frac{1}{t_u}$

$y = \frac{d}{2} (\frac{1}{t_d} - \frac{1}{t_u})$

y = 8 mph
What a great detailed reply! Thank you for your time and effort.

5. ## ok.....

Originally Posted by Greengoblin
$v=\frac{s}{t}$

$x+y=\frac{16}{20}$

$x-y=\frac{16}{30}$

Then you have two simultaneous equations to solve.
I thank you for your time and effort.