Answer:
Approximately [tex]0.478\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
The velocity of the canoe relative to the ground is equal to the vector sum of:
[tex]\begin{aligned}& v(\text{canoe, relative to ground}) \\ =\; & v(\text{canoe, relative to flow}) + v(\text{flow, relative to ground})\end{aligned}[/tex].
Hence, the velocity of the canoe relative to the water can be found by subtracting the velocity of the flow (relative to the ground) from the velocity of the canoe relative to the ground:
[tex]\begin{aligned} & v(\text{canoe, relative to ground}) - v(\text{flow, relative to ground}) \\ =\; & v(\text{canoe, relative to flow})\end{aligned}[/tex].
Let the first component of velocity denote motions the north-south direction, with north being the positive direction. Let the second component denote motions in the east-west direction, with east being the positive direction.
Represent the two given quantities [tex]v(\text{canoe, relative to ground})[/tex] and [tex]v(\text{flow, relative to ground})[/tex] as vectors.
It is given that [tex]v(\text{canoe, relative to ground})[/tex] points in the southwest direction, which would be [tex]135^{\circ}[/tex] from north (positive direction of the first component) and [tex]45^{\circ}[/tex] from east (positive direction of the second component.) Hence, the vector representation of this quantity would be:
[tex]\displaystyle v(\text{canoe, relative to ground}) = \begin{bmatrix}0.300\, \cos(135^{\circ}) \\ 0.300\, \cos(45^{\circ})\end{bmatrix}\; {\rm m\cdot s^{-1}}[/tex].
Similarly, since [tex]v(\text{flow, relative to ground})[/tex] points to the south, this vector would be at [tex]180^{\circ}[/tex] from north and [tex]90^{\circ}[/tex] from east:
[tex]\displaystyle v(\text{flow, relative to ground}) = \begin{bmatrix}0.640\, \cos(180^{\circ}) \\ 0.640\, \cos(90^{\circ})\end{bmatrix}\; {\rm m\cdot s^{-1}}[/tex].
Subtract [tex]v(\text{flow, relative to ground})[/tex] from [tex]v(\text{canoe, relative to ground})[/tex] to find [tex]v(\text{canoe, relative to flow})[/tex]:
[tex]\begin{aligned} & v(\text{canoe, relative to flow}) \\ =\;& v(\text{canoe, relative to ground}) - v(\text{flow, relative to ground}) \\ =\; & \begin{bmatrix}0.300\, \cos(135^{\circ}) \\ 0.300\, \cos(45^{\circ})\end{bmatrix}\; {\rm m\cdot s^{-1}}- \begin{bmatrix}0.640\, \cos(180^{\circ}) \\ 0.640\, \cos(90^{\circ})\end{bmatrix}\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
The magnitude of this vector would be:
[tex]\begin{aligned}& \|v(\text{canoe, relative to flow})\| \\ =\; & \sqrt{(0.300\, \cos(135^{\circ}) - 0.640\, \cos(180^{\circ}))^{2} + (0.300\, \cos(45^{\circ}) - 0.640\, \cos(90^{\circ}))^{2}}\; {\rm m\cdot s^{-1}} \\ \approx\; & 0.478\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
