An ant walks on a piece of graph paper straight along the x axis a distance of 10.0 cm in 2.20 s. It then turns left 30.0 degrees and walks in a straight line another 10.0 cm in 1.90 s. Finally, it turns another 70.0 degrees to the left and walks another 10.0 cm in 1.60 s.

a)Determine the x component of the ant's average velocity.

b)Determine the y component of the ant's average velocity.

c)Determine the magnitude of ant's average velocity.

d)Determine the direction of ant's average velocity.

a) Well, the ant walks straight along the x-axis for a distance of 10.0 cm in 2.20 s. So, the x-component of its average velocity is simply 10.0 cm divided by 2.20 s. So, the x component of the ant's average velocity is approximately 4.55 cm/s.

b) Now, when the ant turns left 30.0 degrees, it starts moving in the y direction. So, its y-component of average velocity is determined by the distance covered and the time taken in the y direction. As it moves 10.0 cm in 1.90 s, the y component of its average velocity is approximately 5.26 cm/s.

c) To determine the magnitude of the ant's average velocity, we can use the pythagorean theorem. The magnitude is the square root of the sum of the squares of the x-component and y-component of the average velocity. So, the magnitude of the average velocity is:

magnitude = √((4.55 cm/s)^2 + (5.26 cm/s)^2)

Using a calculator, the magnitude of the average velocity is approximately 6.97 cm/s.

d) To determine the direction of the ant's average velocity, we can use trigonometry. We can find the angle using the inverse tangent function. The direction is the angle made by the average velocity vector with the x-axis. So, the direction of the ant's average velocity is:

direction = tan^(-1)((5.26 cm/s) / (4.55 cm/s))

Using a calculator, the direction of the ant's average velocity is approximately 48.5 degrees. Therefore, the direction is 48.5 degrees with respect to the positive x-axis.

To solve this problem, we can break down each step of the ant's motion and calculate the x and y components separately.

a) To find the x component of the ant's average velocity, we need to consider the displacement in the x direction and divide it by the total time taken.

The ant moves 10.0 cm straight along the x-axis in 2.20 s and then another 10.0 cm in 1.90 s. Therefore, the total displacement in the x-direction is 10.0 cm + 10.0 cm = 20.0 cm.

The total time taken is 2.20 s + 1.90 s = 4.10 s.

The x component of the average velocity is the displacement in the x direction divided by the total time taken:
x-component = (20.0 cm) / (4.10 s) = 4.88 cm/s.

Therefore, the x component of the ant's average velocity is 4.88 cm/s.

b) To find the y component of the ant's average velocity, we need to consider the displacement in the y direction and divide it by the total time taken.

After turning left 30.0 degrees, the ant moves 10.0 cm to the left in a straight line. This will give us the y component since it is perpendicular to the x-axis.

The total displacement in the y-direction is (10.0 cm) * sin(30.0 degrees) = 5.0 cm.

The total time taken is 4.10 s from part a.

The y component of the average velocity is the displacement in the y direction divided by the total time taken:
y-component = (5.0 cm) / (4.10 s) = 1.22 cm/s.

Therefore, the y component of the ant's average velocity is 1.22 cm/s.

c) To find the magnitude of the ant's average velocity, we can use the Pythagorean theorem:

Magnitude = √(x-component^2 + y-component^2)
Magnitude = √((4.88 cm/s)^2 + (1.22 cm/s)^2)
Magnitude = √(23.74 cm^2/s^2 + 1.49 cm^2/s^2)
Magnitude = √25.23 cm^2/s^2
Magnitude ≈ 5.02 cm/s.

Therefore, the magnitude of the ant's average velocity is approximately 5.02 cm/s.

d) To find the direction of the ant's average velocity, we can calculate the angle it makes with the positive x-axis.

Direction = arctan(y-component / x-component)
Direction = arctan(1.22 cm/s / 4.88 cm/s)
Direction ≈ 14.2 degrees.

Therefore, the direction of the ant's average velocity is approximately 14.2 degrees with respect to the positive x-axis.

To solve this problem, we need to break down the ant's motion into horizontal and vertical components.

a) Determine the x component of the ant's average velocity:
The ant moves along the x-axis initially and turns left twice. However, since the ant's motion is along the x-axis, it means there will be no change in the x component of its velocity throughout the motion. Thus, the x component of the ant's average velocity is simply the displacement along the x-axis divided by the total time taken.

Displacement along x-axis = 10.0 cm
Total time taken = 2.20 s + 1.90 s + 1.60 s = 5.70 s

x component of average velocity = displacement along x-axis / total time taken
= 10.0 cm / 5.70 s

b) Determine the y component of the ant's average velocity:
Since the ant turns left twice, it means there will be a change in the y component of its velocity. To find the y component of the average velocity, we need to calculate the vertical displacement and divide it by the total time taken.

The vertical displacement is given by:
Vertical displacement = (10.0 cm) sin(30.0 degrees) + (10.0 cm) sin(70.0 degrees)

Total time taken = 2.20 s + 1.90 s + 1.60 s = 5.70 s

y component of average velocity = vertical displacement / total time taken

c) Determine the magnitude of the ant's average velocity:
The magnitude of average velocity is given by the square root of the sum of the squares of the x and y components of the average velocity.

Magnitude of average velocity = sqrt((x component)^2 + (y component)^2)

d) Determine the direction of the ant's average velocity:
The direction of the ant's average velocity can be determined using the arctangent of the y component divided by the x component.

Direction of average velocity = arctan(y component / x component)

By following these steps, you can calculate the answers to parts (a), (b), (c), and (d) of the problem.

a. Velocity vectors:

V1 = 10 cm / 2.2 s = 4.55 cm / s @ 0 deg.

V2 = 10 cm / 1.9 s = 5.26 cm / s @ 30 deg.

V3 = 10 cm / 1.6 s = 6.25 cm / s @ 100 deg.

X = hor = 4.65 + 5.26*cos30 + 6.25*cos100 = 4.65 + 4.56 + (-1.08),
X = 8.12 cm / s.

b. Y = ver = 5.26*sin30 + 6.25*sin100,
= 2.63 + 6.16 = 8.79 cm / s.

c. Z^2 = X^2 + Y^2,
= (8.12)^2 + (8.79)^2,
= 65.93 + 77.26,
= 143.19,
Z = sqrt(143.19) = 11.97 cm / s
= Magnitude.

d. tanA = Y / X = 8.79 / 8.12=1.08,
A = 47.3 deg CCW = 42.7deg. E of N.
= Direction.