Question

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?

Answer 1

`d = t_1 xx ( p + c)`
`d = t_2 xx (p-c)`

Explanation:

In the equations above

`d = distance = 560 m`

`t_1 = time` downstream = 70 sec

`t_2 = time` upstream = 80 sec

`p = rate` of the paddlers without the current.

`c = rate` of the current.

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

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

`70( p + c ) = 80( p-c)`

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

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

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

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

`560 = 70 xx ( 15c + c)`

`560 = 70 xx 16c`

`560 = 1120 c` divide both sides by 1120

`560/1120 = 1120c/1120`

`.5 m/s = c` put this value in for c and solve for p

`560 = 80 xx ( p - .5)`

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

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

`600 = 80p` divide both sides by 80

`600/80 = 80/80p`

`75 m/s = p`

Answer 2

`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= 560/70=8` and `V_u=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.