# In a canoe race, a team paddles downstream 560 meters in 70 seconds. The same team makes the trip back upstream to the starting point in 80 seconds. How do you write a system of two equations in two variables that models this problem?

Jul 2, 2017

$d = {t}_{1} \times \left(p + c\right)$
$d = {t}_{2} \times \left(p - c\right)$

#### Explanation:

In the equations above

$d = \mathrm{di} s \tan c e = 560 m$

${t}_{1} = t i m e$ downstream = 70 sec

${t}_{2} = t i m e$ upstream = 80 sec

$p = r a t e$ of the paddlers without the current.

$c = r a t e$ of the current.

$d = d$ so the two equations can be set equal

# T_1 xx (p + c ) = T_2 xx ( p - c)

$70 \left(p + c\right) = 80 \left(p - c\right)$

$70 p + 70 c = 80 p - 80 c$ add 80c and subtract 70p from both sides

$70 p - 70 p + 70 c + 80 c = 80 p - 70 p + - 80 c + 80 c$ equals

$150 c = 10 p$ divide both sides by 10

$15 c = p$ so in place of p 15 c can be substituted

$560 = 70 \times \left(15 c + c\right)$

$560 = 70 \times 16 c$

$560 = 1120 c$ divide both sides by 1120

$\frac{560}{1120} = 1120 \frac{c}{1120}$

$.5 \frac{m}{s} = c$ put this value in for c and solve for p

$560 = 80 \times \left(p - .5\right)$

$560 = 80 p - 40$ add 40 to both sides

$560 + 40 = 80 p - 40 + 40$ equals

$600 = 80 p$ divide both sides by 80

$\frac{600}{80} = \frac{80}{80} p$

$75 \frac{m}{s} = p$

Jul 2, 2017

${V}_{d} = {V}_{c} + {V}_{s}$
${V}_{u} = {V}_{c} - {V}_{s}$

where ${V}_{d}$ is velocity downstream,${V}_{u}$ is velocity upstream,${V}_{c}$ is velocity of Canoe and ${V}_{s}$ is velocity of stream

#### Explanation:

This is a problem involving relative velocity.The canoe moves faster while moving downstream.Let's denote ${V}_{c}$ as Velocity of Canoe,${V}_{s}$ as velocity of stream ,Velocity upstream by ${V}_{u}$ and Velocity downstream by ${V}_{d}$. Since velocity is a vector quantity, the downstream and upstream velocities can be found by adding and subtracting stream velocity from canoe velocity.These can be represented as equations mentioned above.

Additionally ${V}_{d} = \frac{560}{70} = 8$ and ${V}_{u} = \frac{560}{80} = 7$ ,which when substituted gives $8 = {V}_{c} + {V}_{s}$ and $7 = {V}_{c} - {V}_{s}$ .
With these 2 equations values for ${V}_{c}$ and ${V}_{s}$ can be found.