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Problem 22 Medium Difficulty

A space vehicle is coasting at a constant velocity of 21.0 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 0.320 ${m} / {s}^{2}$ in the x direction. After 45.0 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 measured from the y direction.

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Video Transcript

All right. So here, we're gonna do problem number 22. Eso first start by writing down what we know in the questions. So we know V. So we're initial speed in the wind direction is 21 meters per second. We know our acceleration are why is zero meters per second? Because we're on the accelerating the extraction and are therefore initial velocity. And the Y is equal to our I'm lost in the light because we haven't changed our speed at all. Now that's right. Down when we know for X, our ex initial velocity is zero meters per second. Our acceleration in the ex direction is evil to you. 0.3 to 0 meters per second squared and our time that we're accelerating for is 45 seconds. Um, all right, so no, we confined our final X velocity. So that's really big. Find final ex velocity. So we know that from the formula, our final X velocity is equal to our initial X velocity, plus the acceleration times, the time that we're accelerating for so here. We're going to get zero because we're not moving from appear in our initial X direction plus acceleration which is your 0.3 to 0 meters per second squared times 45 seconds. And that's gonna give us a fun of lost in the acts of 14.4 meters per second. Okay, Perfect. So now we have that. So now that we have our final, why velocity and our final X velocity, we can find the magnitude of our velocity. Sonu, starting your weight word here. So the magnitude of los e mag magnitude of velocity is equal to the squares at it. So essentially, the magnitude of the velocity is equal to the square root of the X squared plus v y squared. So this is equal to the square root of four 14 meters per second squared, plus 21 meters per second squared. And this gives us ah, velocity magnitude of 25.5 meters per second. Death perfect. So now we know we have Ah, complete triangle now. So I'm gonna start in on your white board. Our triangle is R v r, which we just found this year. Then we have our y velocity, which we had in the first part. That didn't change. They have our ex velocity s. We found this to be 14.4 meters per second. We had our why to be 21.0 meters per second and we just calculated are our velocity to be 25.5 meters per second along some angle feed from the exact So when you get a question like this, um, now we can use whatever trigger identity we want to find theater. We can use co sign, sign or tangent. I like to use tangent. So we're gonna do that so theta equals inverse tangent. And since we're using, uh, 10 we have to use Ah, the ah opposite side divided by the adjacent sides. So we get 21.0 meters per second over 14.4 meters per second, and this gives us a final angle of 34.4 degrees. Um, but this is east east of north, because we're going to the right of north notice how this angle is larger than this angle. That means we're closer to the north actives than we are to the east axes. So we just go that far east of the north axes. So then we can write our in summary or those summary box here. The magnitude of our velocity was 25.5 meters per second, our angle theta was 34.4 meters per second and our velocity was 25.5 meters per second at 34.4 degrees east of north. So thanks for watching hope that video is hopeful and

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Video Transcript

John D. Cutnell, Kenneth W. Johnson

Physics

Christina Krawiec

Liev Birman

Marshall Styczinski

Zachary Mitchell

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics

Christina Krawiec

Liev Birman

Marshall Styczinski

Zachary Mitchell

John D. Cutnell, Kenneth W. JohnsonPhysics

Christina Krawiec

Liev Birman

Marshall Styczinski

Zachary Mitchell

John D. Cutnell, Kenneth W. Johnson

Physics