A fisherman who can paddle at 4 km/h in still water, heads a canoe due north. The water is moving at a steady speed of 2 km/h due east. Using a diagram, or otherwise, find (a) the magnitude of the resultant velocity of the canoe, (b) the angle of travel of the canoe relative to North.

$$v_R = \sqrt{v_c^2 + v_w^2}$$ $$v_R = \sqrt{(4 \text{ km/h})^2 + (2 \text{ km/h})^2}$$

$$v_R = \sqrt{16 \text{ km}^2/\text{h}^2 + 4 \text{ km}^2/\text{h}^2}$$ $$v_R = \sqrt{20 \text{ km}^2/\text{h}^2}$$ $$v_R = 2\sqrt{5} \text{ km/h}$$ $$v_R \approx 4.47 \text{ km/h}$$

The magnitude of the resultant velocity is $\boxed{\text{4.47 km/h}}$.

b) The angle of travel of the canoe relative to North.

The opposite side is the water current velocity ($v_w$) and the adjacent side is the canoe's velocity ($v_c$).

$$\tan(\theta) = \frac{v_w}{v_c}$$ $$\tan(\theta) = \frac{2 \text{ km/h}}{4 \text{ km/h}}$$ $$\tan(\theta) = 0.5$$

$$\theta = \arctan(0.5)$$ $$\theta \approx 26.57^\circ$$

The angle of travel relative to North is $\boxed{26.57^\circ \text{ East of North}}$.

Question 4:

• Initial velocity ($u$) = 0 m/s (starts at rest)
• Acceleration due to free fall ($g$) = 10 m/s$^2$

a) Calculate the velocity of the fruit 1.0 s after it starts to fall.

$$v = 10 \text{ m/s}$$

The velocity of the fruit after 1.0 s is $\boxed{\text{10 m/s}}$.

b) The fruit takes 2.0 s to hit the ground. Calculate the distance the fruit falls before hitting the ground.

$$s = 0 + \frac{1}{2}(10 \text{ m/s}^2)(4.0 \text{ s}^2)$$ $$s = 5 \text{ m/s}^2 \times 4.0 \text{ s}^2$$ $$s = 20 \text{ m}$$

The distance the fruit falls is $\boxed{\text{20 m}}$.