A rocket-powered hockey puck moves on a horizontal frictionless table. The figure below shows the graphs of vx and vy, the x- and y-components of the puck’s velocity. The puck starts at the origin.

The graph of the X-velocity is v=8t cm/s and the Y-velocity is V=30cm/s

Part A
In which direction is the puck moving at t = 2 s ? Give your answer as an angle from the x-axis.
Express your answer using two significant figures.

Part B
How far from the origin is the puck at 3 s ?
Express your answer to two significant figures and include the appropriate units.

B. X = (1/2)(a)(t)^2 = (1/2)(8)(3)^2 = 36 cm

Y = (v(0))(t) = (30)(3) = 90cm
D = sqrt(x^2+y^2) = sqrt(36^2+90^2) = sqrt(9396)

~ 97 cm

Herny is wrong

Part A: In which direction is the puck moving at t = 2s?

Well, since we have the x-velocity and the y-velocity, we can find the angle by using some trigonometry. Let's call the angle θ.

We know that tan(θ) = vy/vx. Plugging in the given values, we get tan(θ) = 30 cm/s / 8 cm/s.

Using our trusty calculator, we find that tan(θ) ≈ 3.75.

Now, we can find θ by taking the arctan of both sides: θ ≈ arctan(3.75).

Using our calculator again, we find that θ ≈ 75.96°.

So, at t = 2s, the puck is moving at an angle of approximately 75.96° from the x-axis.

Part B: How far from the origin is the puck at t = 3s?

To find the distance from the origin, we can use the formula d = √(x^2 + y^2), where x and y are the x and y coordinates of the puck at t = 3s.

From the given information, we have:

x = vx * t = 8 cm/s * 3s = 24 cm
y = vy * t = 30 cm/s * 3s = 90 cm

Plugging these values into our formula, we have:

d = √(24^2 + 90^2) ≈ √(576 + 8100) ≈ √8676 ≈ 93.1 cm

So, at t = 3s, the puck is approximately 93.1 cm away from the origin.

Part A:

To determine the direction in which the puck is moving at t = 2s, we need to find the angle between the x-axis and the velocity vector of the puck at t = 2s.

From the given information, we know that the x-component of the puck's velocity, vx, is given by the equation vx = 8t cm/s. Plugging in t = 2s, we can find the x-component of the velocity:
vx = 8(2) = 16 cm/s

Similarly, the y-component of the puck's velocity, vy, is given by the equation vy = 30 cm/s.

To find the angle, we can use trigonometry. The angle, θ, can be calculated as:
θ = arctan(vy/vx)
θ = arctan(30/16)

Using a scientific calculator, we can find that θ ≈ 61.3 degrees.

Therefore, the puck is moving at an angle of approximately 61.3 degrees from the x-axis at t = 2s.

Part B:
To find how far from the origin the puck is at t = 3s, we need to calculate the displacement of the puck from the origin.

Since the puck is moving on a frictionless table, its velocity remains constant over time. Therefore, we can use the given velocities to find the displacement.

The x-component of velocity, vx = 8t cm/s, at t = 3s is:
vx = 8(3) = 24 cm/s

The y-component of velocity, vy = 30 cm/s, remains constant.

To find the displacement, we can use the equations of motion. The displacement in the x-direction, dx, can be calculated as:
dx = vx * t
dx = 24 cm/s * 3 s
dx = 72 cm

Similarly, the displacement in the y-direction, dy, can be calculated as:
dy = vy * t
dy = 30 cm/s * 3 s
dy = 90 cm

Since the puck started at the origin, its initial position coordinates are (0, 0). Therefore, the displacement from the origin can be found using the Pythagorean theorem:
displacement = √(dx^2 + dy^2)
displacement = √(72^2 + 90^2)

Using a scientific calculator, we can find that the displacement from the origin is approximately 115.1 cm.

Therefore, the puck is approximately 115.1 cm away from the origin at t = 3s.

i dont kow

115.256 cm

A. X = 8*2 = 16 cm/s. Y = 30 cm/s.

Tan A = Y/X = 30/16 = 1.875.
A = 61.9o

B. X = 8*3 = 24 cm/s. Y = 30 cm/s.
Dx = 24cm/s * 3s = 72 cm.
Dy = 30cm/s * 3s = 90 cm.

D = sqrt(Dx^2 + Dy^2) =