V(hor) = 0.228m/s^2 * 42.3s = 9.64m/s.
a. V^2 + Y^2=(9.64)^2 + (24.7)^2 = 703,
V = sqrt(703) = 26.5m/s = Magnitude.
b. tanA = Y/X = 24.7 / 9.64 = 2.5622,
A = 68.7Deg CCW = 90 - 68.7 = 21.3Deg
East of +Y = Direction.
a. V^2 + Y^2=(9.64)^2 + (24.7)^2 = 703,
V = sqrt(703) = 26.5m/s = Magnitude.
b. tanA = Y/X = 24.7 / 9.64 = 2.5622,
A = 68.7Deg CCW = 90 - 68.7 = 21.3Deg
East of +Y = Direction.
Step 1: Determine the initial velocity components of the vehicle.
Given:
Initial velocity in the y-direction (Vy) = 24.7 m/s (given in the problem)
Initial velocity in the x-direction (Vx) = 0 m/s (there is no initial velocity in the x-direction)
Step 2: Determine the acceleration components of the vehicle during the RCS thruster's operation.
Given:
Acceleration in the x-direction (Ax) = 0.228 m/s^2 (given in the problem)
Acceleration in the y-direction (Ay) = 0 m/s^2 (since there is no acceleration in the y-direction)
Step 3: Determine the time when the RCS thruster is turned off.
Given:
Time during RCS thruster operation (t) = 42.3 s (given in the problem)
Step 4: Calculate the final velocity components of the vehicle.
To calculate the final velocity components, we will use the following equations of motion:
Vx = V0x + Ax * t
Vy = V0y + Ay * t
Considering initial velocity in the x-direction (V0x) is 0 m/s and acceleration in the y-direction (Ay) is 0 m/s^2, the equations simplify to:
Vx = 0 + 0.228 * 42.3
Vy = 24.7 + 0 * 42.3
Calculating Vx:
Vx = 0.228 * 42.3
= 9.6464 m/s
Calculating Vy:
Vy = 24.7
Hence, the final velocity components of the vehicle relative to the space station are:
Vx = 9.6464 m/s
Vy = 24.7 m/s
Step 5: Calculate the magnitude and direction of the final velocity.
To find the magnitude of the final velocity (Vf), we can use the Pythagorean theorem:
Vf = sqrt(Vx^2 + Vy^2)
= sqrt((9.6464)^2 + (24.7)^2)
= sqrt(93.13924416 + 608.09)
= sqrt(701.22924416)
= 26.49 m/s (rounded to two decimal places)
To find the direction of the final velocity, we can use trigonometry. The direction is measured as an angle (θ) in degrees from the +y direction. The angle (θ) can be calculated using the inverse tangent function:
θ = tan^(-1)(Vx / Vy)
= tan^(-1)(9.6464 / 24.7)
= tan^(-1)(0.39119)
= 21.94 degrees (rounded to two decimal places)
Hence, the magnitude of the vehicle's velocity relative to the space station after the RCS thruster is turned off is 26.49 m/s, and its direction is 21.94 degrees from the +y direction.
1. Determine the velocity of the vehicle right after the RCS thruster is turned off.
2. Determine the magnitude and direction of the final velocity.
Let's start with step 1:
1. Determine the velocity of the vehicle right after the RCS thruster is turned off:
We know that the initial velocity in the +y direction is 24.7 m/s, and the vehicle undergoes acceleration in the +x direction at a rate of 0.228 m/s^2 for 42.3 seconds. To find the final velocity in the +x direction, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Using this equation, we have:
v = 0 + (0.228 m/s^2) × (42.3 s)
v ≈ 9.64 m/s
Therefore, the velocity of the vehicle right after the RCS thruster is turned off is approximately 9.64 m/s in the +x direction.
Now let's move to step 2:
2. Determine the magnitude and direction of the final velocity:
To find the magnitude of the final velocity, we can use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.
In this case, the final velocity is the hypotenuse, and the components in the x and y directions are the other two sides. Given that the velocity in the x direction is 9.64 m/s and the velocity in the y direction is 24.7 m/s, we can write:
v^2 = vx^2 + vy^2
Using this equation, we have:
v^2 = (9.64 m/s)^2 + (24.7 m/s)^2
Calculating the square root of both sides, we get:
v ≈ √((9.64 m/s)^2 + (24.7 m/s)^2)
v ≈ 26.6 m/s
Therefore, the magnitude of the final velocity is approximately 26.6 m/s.
To find the direction of the final velocity, we can use trigonometry. The angle (θ) measured from the +y direction can be calculated using the inverse tangent (arctan) function:
θ = arctan(vx / vy)
Substituting the values, we have:
θ = arctan(9.64 m/s / 24.7 m/s)
Calculating this value using a calculator, we get:
θ ≈ 21.6 degrees
Therefore, the direction of the vehicle's velocity relative to the space station, measured from the +y direction, is approximately 21.6 degrees.