A 45 N weight is projected downward with an initial speed of 4.6 m/s. If the air resistance is proportional to the speed and if the limiting speed is double the initial speed, how far has the body traveled when it has reached a speed of 6 m/s?

This problem can be solved using the concept of terminal velocity and the equation of motion under constant acceleration.

1. First, we need to find the acceleration of the weight. Since the limiting speed is double the initial speed, we know that the acceleration due to air resistance is half of the acceleration due to gravity. This is because when an object reaches its terminal velocity, the force due to gravity is balanced by the force due to air resistance.

2. The acceleration due to gravity is 9.8 m/s^2, so the acceleration due to air resistance is 9.8/2 = 4.9 m/s^2.

3. The net acceleration of the weight is the difference between the acceleration due to gravity and the acceleration due to air resistance, which is 9.8 - 4.9 = 4.9 m/s^2.

4. Now we can use the equation of motion v = u + at to find the time it takes for the weight to reach a speed of 6 m/s. Here, v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

5. Plugging in the values, we get 6 = 4.6 + 4.9t. Solving for t, we get t = (6 - 4.6) / 4.9 = 0.2857 seconds.

6. Finally, we can use the equation of motion s = ut + 0.5at^2 to find the distance traveled by the weight. Here, s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

7. Plugging in the values, we get s = 4.6 * 0.2857 + 0.5 * 4.9 * (0.2857)^2 = 1.3143 + 0.2003 = 1.5146 meters.

So, the weight has traveled approximately 1.51 meters when it has reached a speed of 6 m/s.