A turtle ambles leisurely–as turtles tend to do–when it moves from a location with position vector r1,x = 1.49 m and r1,y = -2.83 m in a lettuce garden to another location, with position vector r2,x = 3.17 m and r2,y = -4.49 m, where the lettuce appears to be tastier (but really isn\'t–this turtle can\'t see too well). The excursion takes 325 seconds to complete. What are the components and the magnitude of the turtle\'s average velocity, in meters per second?
X = 3.17 - 1.49 = 1.68 m.
Y = -4.49 - (-2.83) = -1.66 m.
d = Sqrt(X^2+Y^2) = Sqrt(1.68^2+1.6^2) = 2.36 m.
V = d/t = 2.36m/325s = 0.00727m/s.
To find the components and magnitude of the turtle's average velocity, we need to divide the displacement vector by the time taken.
The displacement vector, Δr, can be found by subtracting the initial position vector, r1, from the final position vector, r2:
Δr = r2 - r1
To find the components of Δr, we subtract the corresponding components of the initial and final position vectors:
Δr,x = r2,x - r1,x
Δr,y = r2,y - r1,y
Substituting the given values, we have:
Δr,x = 3.17 m - 1.49 m = 1.68 m
Δr,y = -4.49 m - (-2.83 m) = -1.66 m
Now, to find the magnitude of Δr, we use the Pythagorean theorem:
|Δr| = √(Δr,x^2 + Δr,y^2)
Substituting the values, we have:
|Δr| = √(1.68^2 + (-1.66)^2) = √(2.8224 + 2.7556) = √5.578 = 2.36 m
To find the average velocity, v, we divide the displacement vector by the time:
v = Δr / t
Substituting the values, we have:
v = 2.36 m / 325 s ≈ 0.0073 m/s
Therefore, the components of the turtle's average velocity are Δr,x = 1.68 m and Δr,y = -1.66 m, and the magnitude of the average velocity is approximately 0.0073 m/s.