A space vehicle is coasting at a constant velocity of 24.7 m/s in the +y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.228 m/s2 in the +x direction. After 42.3 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the +y direction.
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.
CORRECTION: a. X^2 + Y^2 =
To find the magnitude and direction of the vehicle's velocity relative to the space station after the RCS thruster is turned off, we can use the following steps:
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.
To find the magnitude and direction of the vehicle's velocity relative to the space station after the RCS thruster is turned off, we can break down the problem into two parts:
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.