A space vehicle is coasting at a constant velocity of 21.5 m/s in the + y direct

Question

A space vehicle is coasting at a constant velocity of 21.5 m/s in the +y direction relative to aspace station. The pilot of the vehicle fires a RCS (reactioncontrol system) thruster, which causes it to accelerate at0.310 m/s2 in the +xdirection. After 44.0 s, the pilot shutsoff the RCS thruster. After the RCS thruster is turned off, findthe following quantities. (a) the magnitude of the vehicle's velocity
1 m/s

(b) the direction of the vehicle's velocity relative to the spacestation Express the direction as an angle measured from the+y direction.
2° to the rightof the +y direction (a) the magnitude of the vehicle's velocity
1 m/s

(b) the direction of the vehicle's velocity relative to the spacestation Express the direction as an angle measured from the+y direction.
2° to the rightof the +y direction

Explanation / Answer

Given in the equation: initial velocity in the y direction: vyo = 21.3m/s acceleration in the x direction:    ax =.310 m/s2 time the thruster is on: 44.0 s in the y direction, acceleration is 0 always, because vyis constant. Therefore: vy(t) =vyo                     denoting vy of time t in the x direction, acceleration is .310 m/s2 andvxo = 0 m/s. Therefore: vx (t) =.310(t)                denoting vx of time t Now insert the time acceleration is on, 44.0s: vx (44.0s) = .310 (44.0) vx (44.0s) ˜ 13.64m/s Now form your vector addition, making a right triangle,21.3m/s in the y direction and13.64m/s in the x direction. Evaluating the Pythagorean Theorem, (21.3)2 +(13.64)2 = (Vfinal)2       --> Solve forVfinal Vfinal ˜ 25.29 m/s Now solve for your angle right of the y-axis, usingtrig. Tan() = opposite/adjacent Tan() = 13.64/21.3 = Tan-1 (13.64/21.3) ˜ 32.6o

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