Question

Interactive Learningware 3.1 reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking 1.40 km due east, 1.63 km due south, and 3.26 km in a direction 78.6 ° north of west. The time required for this trip is 2.016 h. Determine the magnitudes of the bird watcher's (a) displacement and (b) average velocity.

Answer 1

Answer:

(a) The magnitude of the bird watcher's displacement is: 1.739 Km and (b) Average Velocity is: 0.863 Km/h.

Explanation:

To know the magnited of the bird watcher's displacement, we need to use cartesian coordinates. The distance that the bird watcher walked is the vector addition of all movement from the start point. So East = 1.4 - 3.26*Cos(78.6) = 0.756 Km and North = -1.63 + 3.26*Sen (78.6) = 1.566 Km when we have got result of both components axles, we use the pythagorean theorem to get the displacement that the bird watcher walked; R^{2}=x^{2}+y^{2} =\sqrt{0.756^{2}+(1.566)^{2}}=1.739(Km). Then we are going to calculate average velocity as:$V_{Prom} =\frac{delta_x}{delta_t}$ where delta_x is the displacement traveled by the bird watcher and delta_t is the total time using during the movement, so $V_{Prom}=\frac{1.739}{2.016}=0.863(Km/h)$.