A space shuttle is approaching the Alpha International Space Station at a velocity of 12 m/s relative to the space station. A landing cable is fired toward Alpha with a velocity of 3.0 m/s, at an angle of 25 degrees relative to the shuttle. What velocity will the cable appear to have to an observer looking out a window in the space station?
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If we make the direction of the shuttle's motion the x-axis, then the velocity of the cable is
12 m/s in the x direction + 3 m/s at 25 degrees off of the x-axis.
= 12 m/s in the x direction + 3cos(25) m/s at in the x direction + 3sin(25) m/s in the y-direction.
The magnitude of this vector is
√((12 + 3cos(25))^2 + (3sin(25))^2)
= 14.77 m/s
(Rounding to significant figures, the situation is indistinguishable from the cable being fired straight ahead)
12 m/s in the x direction + 3 m/s at 25 degrees off of the x-axis.
= 12 m/s in the x direction + 3cos(25) m/s at in the x direction + 3sin(25) m/s in the y-direction.
The magnitude of this vector is
√((12 + 3cos(25))^2 + (3sin(25))^2)
= 14.77 m/s
(Rounding to significant figures, the situation is indistinguishable from the cable being fired straight ahead)
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So the cable is coming directly toward the observer at the station. However, the shuttle is coming towards at the station at an 25 degree angles. So therefore, 12cos(angle) m/s + 3 m/s would be the correct answer.
Ans = 13.88 m/s
Ans = 13.88 m/s