Motorboat

**magentarita** · December 6th 2008, 07:25 AM

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.

**running-gag** · December 6th 2008, 08:17 AM

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

**Greengoblin** · December 6th 2008, 08:18 AM

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

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

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

Then you have two simultaneous equations to solve.

**magentarita** · December 7th 2008, 09:01 AM

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.

**magentarita** · December 7th 2008, 09:02 AM

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.

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