A skateboarder travels on a horizontal surface with an initial velocity of 4.4 m/s toward the south and a constant acceleration of 2.0 m/s2 toward the east. Let the x direction be eastward and the y direction be northward, and let the skateboarder be at the origin at t=0.

What is her x position at t=0.90s?
What is her y position at t=0.90s?
What is her x velocity component at t=0.90s?
What is her y velocity component at t=0.90s?
Please list steps on how to do them if possible!

all positions are vectors in x,y (east, North) direction.

p(t)=pxo + pyo + Vox*t +Voy*t + 1/2 (ax*t)+1/2 (ay*t)
px(t)=0+0*t+1/2 *2*t
py(t)=0+(-4.4)t
so you put in the value for time t, and you have x and y positions
velocity? Vx=ax*t=2t
Vy=-4.4t (or -y, or South direction)

Oh, how amusing! The skateboarder seems to be a bit confused with their directions. Let's help them out, shall we?

To solve these problems, we'll use the equations of motion in two dimensions. These equations are:

x = x_0 + v_x0 * t + (1/2) * a_x * t^2 (Equation 1)
y = y_0 + v_y0 * t + (1/2) * a_y * t^2 (Equation 2)
v_x = v_x0 + a_x * t (Equation 3)
v_y = v_y0 + a_y * t (Equation 4)

Now, let's calculate the values step by step:

1. x position at t = 0.90s:
Since the skateboarder travels towards the south, which is the negative y direction, there is no x component in her initial velocity. As a result, her x position at t = 0.90s will be the same as her initial x position, which is 0 units.

2. y position at t = 0.90s:
Using Equation 2, we have:
y = y_0 + v_y0 * t + (1/2) * a_y * t^2
Plugging in the given values:
y = 0 + 4.4 * 0.90 + (1/2) * 2.0 * (0.90)^2

Solve this expression, and you'll find the y position at t = 0.90s.

3. x velocity component at t = 0.90s:
Since the skateboarder has a constant acceleration towards the east (positive x direction), her x velocity component increases at a rate of 2.0 m/s^2. Using Equation 3, we have:
v_x = v_x0 + a_x * t

Plugging in the given values:
v_x = 0 + 2.0 * 0.90

Solve this expression to find her x velocity component at t = 0.90s.

4. y velocity component at t = 0.90s:
The skateboarder starts with an initial velocity towards the south (negative y direction), but since there is no acceleration in the y direction, her y velocity component remains constant. So, her y velocity component at t = 0.90s will be the same as her initial velocity, which is 4.4 m/s towards the south.

Now, go ahead and crunch those numbers to find out the actual values!

To solve this problem step-by-step, we can break it down into four parts: finding the x position, the y position, the x velocity component, and the y velocity component at t=0.90s.

1. Finding the x position at t=0.90s:
The skateboarder has a constant acceleration in the x direction (eastward), but no initial velocity in that direction. We can use the equation:
x = x0 + v0t + (1/2)at^2
where x0 is the initial position (0 since the skateboarder is at the origin), v0 is the initial velocity (0 since there is no initial velocity in the x direction), a is the acceleration (2.0 m/s^2), and t is the time (0.90s). Plugging in the values, we have:
x = 0 + 0(0.90) + (1/2)(2.0)(0.90)^2
x = (1/2)(2.0)(0.81)
x = 0.81m

Therefore, the skateboarder's x position at t=0.90s is 0.81m.

2. Finding the y position at t=0.90s:
Since the skateboarder is initially traveling toward the south (negative y direction), we can use the equation:
y = y0 + v0t + (1/2)at^2
where y0 is the initial position (0 since the skateboarder is at the origin), v0 is the initial velocity (-4.4 m/s), a is the acceleration (0 since there is no acceleration in the y direction), and t is the time (0.90s). Plugging in the values, we have:
y = 0 + (-4.4)(0.90) + (1/2)(0)(0.90)^2
y = -3.96m

Therefore, the skateboarder's y position at t=0.90s is -3.96m.

3. Finding the x velocity component at t=0.90s:
The x velocity component can be found using the equation:
vx = v0x + at
where v0x is the initial velocity in the x direction (0 since there is no initial velocity in the x direction), a is the acceleration (2.0 m/s^2), and t is the time (0.90s). Plugging in the values, we have:
vx = 0 + (2.0)(0.90)
vx = 1.80 m/s

Therefore, the skateboarder's x velocity component at t=0.90s is 1.80 m/s.

4. Finding the y velocity component at t=0.90s:
The y velocity component can be found using the equation:
vy = v0y + at
where v0y is the initial velocity in the y direction (-4.4 m/s), a is the acceleration (0 since there is no acceleration in the y direction), and t is the time (0.90s). Plugging in the values, we have:
vy = (-4.4) + (0)(0.90)
vy = -4.4 m/s

Therefore, the skateboarder's y velocity component at t=0.90s is -4.4 m/s.

To solve these questions, we can break down the motion into its x and y components. We'll use the kinematic equations of motion to find the position and velocity components at a given time. Let's solve each question step by step:

1. What is her x-position at t = 0.90 s?

To find her x-position, we need to determine the displacement in the x-direction. Since the skateboarder starts with an initial velocity towards the south (opposite to the x-axis), her displacement in the x-direction will only occur due to acceleration.

We can use the formula:
x = x0 + v0x * t + (1/2) * ax * t^2

Here, x0 represents the initial x-position (which is 0), v0x is the initial x-velocity (0 since she has no initial velocity in the x-direction), ax is the acceleration in the x-direction (2.0 m/s^2), and t is the time (0.90 s).

Plugging in the values, we get:
x = 0 + 0 * 0.90 + (1/2) * 2.0 * (0.90)^2

Simplifying the equation gives us:
x = (1/2) * 2.0 * 0.81
x = 0.81 m

Therefore, her x-position at t = 0.90 s is 0.81 m to the east.

2. What is her y-position at t = 0.90 s?

To find her y-position, we again use the formula for displacement, but this time in the y-direction. Since the skateboarder is initially at the origin in the y-direction, her displacement will only occur due to her initial velocity in the y-direction.

The formula becomes:
y = y0 + v0y * t + (1/2) * ay * t^2

Here, y0 represents the initial y-position (which is 0), v0y is the initial y-velocity (4.4 m/s since she has a velocity towards the south), ay is the acceleration in the y-direction (0 since there's no acceleration in the y-direction), and t is the time (0.90 s).

Plugging in the values, we get:
y = 0 + 4.4 * 0.90 + (1/2) * 0 * (0.90)^2

Simplifying the equation gives us:
y = 3.96 m

Therefore, her y-position at t = 0.90 s is 3.96 m to the north.

3. What is her x-velocity component at t = 0.90 s?

The x-velocity component represents her velocity in the x-direction at a given time. Since there's a constant eastward acceleration of 2.0 m/s^2 in the x-direction, her x-velocity will increase linearly with time.

The formula for x-velocity is:
vx = v0x + ax * t

Here, v0x represents the initial x-velocity (0 since she has no initial velocity in the x-direction), ax is the acceleration in the x-direction (2.0 m/s^2), and t is the time (0.90 s).

Plugging in the values, we get:
vx = 0 + 2.0 * 0.90

Simplifying the equation gives us:
vx = 1.80 m/s

Therefore, her x-velocity component at t = 0.90 s is 1.80 m/s to the east.

4. What is her y-velocity component at t = 0.90 s?

The y-velocity component represents her velocity in the y-direction at a given time. Since she starts with an initial velocity towards the south, her y-velocity will remain constant as there is no acceleration in the y-direction.

The formula for y-velocity is:
vy = v0y

Here, v0y represents the initial y-velocity (4.4 m/s since she has a velocity towards the south).

Therefore, her y-velocity component at t = 0.90 s remains constant at 4.4 m/s downwards.

In summary:
- Her x-position at t = 0.90 s is 0.81 m to the east.
- Her y-position at t = 0.90 s is 3.96 m to the north.
- Her x-velocity component at t = 0.90 s is 1.80 m/s to the east.
- Her y-velocity component at t = 0.90 s is 4.4 m/s downwards.