A space vehicle is coasting at a constant velocity of 20.7 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 0.260 m/s2 in the +x direction. After 42.0 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find the following quantities.
(a) the magnitude of the vehicle's velocity
m/s
(b) the direction of the vehicle's velocity relative to the space station Express the direction as an angle measured from the +y direction.
° to the right of the +y direction
X = hor. = at = 0.260m/s^2 * 42s = 10.92m/s.
Y = ver. = 20.7m/s.
a. V = sqrt(X^2 + Y^2),
V = sqrt((10.92)^2 + (20.7)^2) = 23.4m/s.
b. tanA = Y / X = 20.7 / 10.92=1.8956.
A = 62.2 deg.,N of E.
A = 90 - 62.2 = 27.8 deg.,E of N.
Where are you getting 62.2 deg
you take the inverse tan. So tan-1(1.8956)= 62.2
To find the magnitude of the vehicle's velocity after the RCS thruster is turned off, we need to calculate the final velocity.
Initially, the vehicle was coasting at a constant velocity of 20.7 m/s in the +y direction relative to the space station. Now, when the RCS thruster is turned on, the vehicle accelerates at 0.260 m/s² in the +x direction for a duration of 42.0 seconds.
To calculate the final velocity, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
Initial velocity (u) = 20.7 m/s (_y)
Acceleration (a) = 0.260 m/s² (_x)
Time (t) = 42.0 s
Using the equation, we can calculate the final velocity in the x-direction (v_x) and y-direction (v_y).
v_x = u_x + a_x * t
= 0 + 0.260 * 42.0
= 10.92 m/s (_x)
v_y = u_y + a_y * t
= 20.7 + 0 * 42.0
= 20.7 m/s (_y)
The magnitude of the final velocity (v) can be calculated using the Pythagorean theorem:
v = √(v_x² + v_y²)
= √(10.92² + 20.7²)
≈ 23.4 m/s
Therefore, the magnitude of the vehicle's velocity after the RCS thruster is turned off is approximately 23.4 m/s.
To calculate the direction of the vehicle's velocity relative to the space station, we need to find the angle (θ) measured from the +y direction.
θ = arctan(v_x / v_y)
θ = arctan(10.92 / 20.7)
≈ 29.6°
Thus, the direction of the vehicle's velocity relative to the space station is approximately 29.6° to the right of the +y direction.