- #1

Mohammed17

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**1. At a time when mining asteroids has become feasible, astronauts have connected a line between their 3500-kg space tug and a 6200-kg asteroid. Using their tug's engine, they pull on the asteroid with a force of 490 N. Initially the tug and the asteroid are at rest, 450 m apart. How much time does it take for the tug and the asteroid to meet?**

G:

Ok so I know that the 3500 kg space tug (exerting 490 N on the asteroid) will also feel a force exerted on it by the 6200 kg asteroid (Newton's third law) of -490 N (or a pull of 490 N on the itself by the asteroid).

I also know that the distance is 450 meters between the asteroid and the space-tug. This will help me later on in the kinematic portion.

I know that Vo (Initial Velocity) is 0 for both. I also know that they both will experience a velocity towards each other once they start to tug on the asteroid.

I don't know whether to use Sigma Fx = m*a

and go from there... I tried that but in the end I have two variables: aST and aAsteroid

or should I just say that 490 N = Mass of ST*acceleration of ST

ad 490 N = Mass of Asteroid*acceleration of Asteroid

So you get:

490 N / 3500 kg = 0.14 m/s^2 = Acceleration of spacetug

and

490 N / 6200 kg = 0.079 m/s^2 = Acceleration of Asteroid\

Then where do I go from there?

Do I plug it into:

d = Vot + 1/2at^2 I know Vo = 0

450 m = 0 + 1/2at^2

450 m = 1/2(Acceleration of Space Tug + Acceleration of Asteroid)^2

SquareRoot of [(2 x 450 m)/(0.14 + 0.079)] = t

**I got t = 64.106 seconds**

THerefore, they will meet each other after 64.11 seconds. Is that correct?

THerefore, they will meet each other after 64.11 seconds. Is that correct?

Please help. Thank you.

- Mohammed.