A turtle ambles leisurely–as turtles tend to do–when it moves from a location with position vector r1,x = 1.53 m and r1,y = -2.25 m in a lettuce garden to another location, with position vector r2,x = 3.91 m and r2,y = -4.65 m, where the lettuce appears to be tastier (but really isn\'t–this turtle can\'t see too well). The excursion takes 5.99 minutes to complete. What are the components and the magnitude of the turtle's average velocity, in meters per second?

r2-r1 = ((3.91-1.53),(-4.65+2.25))

= (2.38,-2.40)

total travel: √(2.38^2 + 2.40^2) = 3.38m

avg speed = distance/time = 3.38m/5.99min * 1min/60s = 0.0094 m/s

thank you!! =)

Why did the turtle cross the garden? To get to the tastier lettuce... even though it can't even see well! Anyway, let's calculate its average velocity.

The change in position, or displacement, is given by the difference between the final position and initial position vectors:
Δr,x = r2,x - r1,x = 3.91 m - 1.53 m = 2.38 m
Δr,y = r2,y - r1,y = -4.65 m - (-2.25 m) = -2.40 m

The elapsed time is 5.99 minutes, but we want the velocity in meters per second, so let's convert minutes to seconds:
Δt = 5.99 min * (60 s / 1 min) = 359.4 s

Now we can calculate the components of average velocity:
v,x = Δr,x / Δt = 2.38 m / 359.4 s ≈ 0.00663 m/s
v,y = Δr,y / Δt = -2.40 m / 359.4 s ≈ -0.00668 m/s

The magnitude of the average velocity is given by:
|v| = √(v,x^2 + v,y^2) = √((0.00663 m/s)^2 + (-0.00668 m/s)^2) ≈ 0.00938 m/s

So, the turtle's average velocity is approximately 0.00663 m/s in the x-direction and -0.00668 m/s in the y-direction. The magnitude of its average velocity is approximately 0.00938 m/s.

To find the components and magnitude of the turtle's average velocity, we can use the formula:

Average Velocity = (Change in Position) / (Change in Time)

To find the change in position, we subtract the initial position vector (r1) from the final position vector (r2).

Change in Position = r2 - r1

Let's calculate the change in position first:

Change in Position (Δr) = (r2,x - r1,x) î + (r2,y - r1,y) ĵ

Plugging in the given values:

Δr = (3.91 m - 1.53 m) î + (-4.65 m - (-2.25 m)) ĵ
= 2.38 m î -2.40 m ĵ

Next, we need to calculate the magnitude of the change in position, which is the magnitude of the average velocity.

Magnitude of the Change in Position (|Δr|) = √((Δr,x)² + (Δr,y)²)

Plugging in the values of Δr, we get:

|Δr| = √((2.38 m)² + (-2.40 m)²)
= √(5.6644 m² + 5.76 m²)
= √(11.4244 m²)
= 3.38 m

So, the magnitude of the turtle's average velocity is 3.38 m.

To find the components of the average velocity, we divide the change in position by the change in time:

Average Velocity = Δr / Δt

Plugging in the given time (5.99 minutes = 5.99 * 60 seconds) and the values of Δr:

Average Velocity = (2.38 m / 359.4 s) î + (-2.40 m / 359.4 s) ĵ

Simplifying, we find:

Average Velocity = 0.0066 m/s î - 0.0067 m/s ĵ

Therefore, the components of the turtle's average velocity are approximately 0.0066 m/s in the x-direction and -0.0067 m/s in the y-direction.