2. during 4.0 min on a lake, a loon moves 25.0 m e 30.0° n and then 75.0 m e 45.0° s. determine the loons…

Explanation:

Step1: Calculate total distance

The loon moves two - part distances. The total distance $d$ is the sum of the magnitudes of the two displacements. So $d = 250+750=1000$ m.

Step2: Resolve displacements into components

For the first displacement $\vec{d}1 = 250$ m [E 30.0° N], the x - component $d{1x}=250\sin30^{\circ}=125$ m and the y - component $d_{1y}=250\cos30^{\circ}=250\times\frac{\sqrt{3}}{2}\approx216.5$ m. For the second displacement $\vec{d}2 = 750$ m [E 45.0° S], the x - component $d{2x}=750\sin45^{\circ}=750\times\frac{\sqrt{2}}{2}\approx530.3$ m and the y - component $d_{2y}=- 750\cos45^{\circ}=-750\times\frac{\sqrt{2}}{2}\approx - 530.3$ m.

Step3: Calculate total x and y components of displacement

The total x - component of displacement $D_x=d_{1x}+d_{2x}=125 + 530.3=655.3$ m. The total y - component of displacement $D_y=d_{1y}+d_{2y}=216.5-530.3=-313.8$ m. Then the magnitude of the displacement $\vec{D}=\sqrt{D_x^{2}+D_y^{2}}=\sqrt{(655.3)^{2}+(-313.8)^{2}}\approx727$ m.

Step4: Calculate average speed

The time $t = 4.0$ min $=4\times60 = 240$ s. The average speed $v_{avg}=\frac{d}{t}=\frac{1000}{240}\approx4.17$ m/s.

Step5: Calculate average velocity

The average velocity $\vec{v}{avg}=\frac{\vec{D}}{t}$. The magnitude of the average velocity $v{avg - vel}=\frac{D}{t}=\frac{727}{240}\approx3.03$ m/s.

Answer:

(a) Total distance travelled: 1000 m (b) Total displacement: approximately 727 m (c) Average speed: approximately 4.17 m/s (d) Average velocity: approximately 3.03 m/s