# Relative velocity of a canoe

1. Jul 10, 2007

### Edwardo_Elric

1. The problem statement, all variables and given/known data
A canoe has a velocity of 0.30m/s northwest relative to the earth. The canoe is on a river that is flowing 0.50m/s west relative to the earth. Find the velocity (magnitude and direction ) of the canoe relative to the river.

2. Relevant equations
E = earth
R = river
C = canoe
$$V_{C/E} = V_{C/R} + V_{R/E}$$
$$V_{C/E} - V_{R/E} = V_{C/R}$$

3. The attempt at a solution
the answer at the back of the book is .36m/s 53.6 degrees east of north

to get to the book's answer, i used the pythagorean theorem
$$\sqrt{{V_{C/E}}^2 - {V_{R/E}}^2} = V_{C/R}$$
and i confusingly substituted
\sqrt{((0.30)cos(45) - .50)^2 - ((.30) sin 45))^2} = V_{C/R}
then get .36
and i dont know how the V_{C/E}'s x components got involved ....

2. Jul 10, 2007

### Edwardo_Elric

ohhh
since : V_{C/E} - V_{R/E} = V_{C/R}
x component: -(.21)- (-.50)
y component: .21 - 0
(.29)^2 + (.21)^2 = V_resultant^2
0.36

$$\theta = arctan\frac{.21}{.29}$$
$$\theta = 35.9$$degres north of east
or 90 - 35.9 = 54.1 e of n

3. Jul 10, 2007

### Staff: Mentor

I like these kind of threads!