Finding the velocity and covered distance of Astronaut pushing asteroid

In summary, the asteroid is going 3.65m/s in the opposite direction of the astronaut after a 5 second constant velocity phase.
  • #1
soul_observer
10
0

Homework Statement


An astronaut of mass 60 kg and a small asteroid of mass 40 are initially at rest with respect to the space station. The astronaut pushes the asteroid with a constant force of magnitude 250N for 0.35s. Gravitational forces are negligible. A)How far apart are the astronaut and asteroid 5.00s after the astronaut stops pushing? B) What is their relative speed at this time?


Homework Equations


TotalForcex=ma
Delta x=(1/2)(delta v)(delta t)
Delta v=a(delta t)


The Attempt at a Solution


1. Total force of x is 250N, so 250N=ma=250N/40= acceleration = 6.25m/s^2
2. (deltax)=(1/2)(6.25)(5.35)^2 = 89.4
3. Vf=(6.25)(5.35)=33.4 m/s

Does this look alright?
 
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  • #2
soul_observer said:

Homework Statement


An astronaut of mass 60 kg and a small asteroid of mass 40 are initially at rest with respect to the space station. The astronaut pushes the asteroid with a constant force of magnitude 250N for 0.35s. Gravitational forces are negligible. A)How far apart are the astronaut and asteroid 5.00s after the astronaut stops pushing? B) What is their relative speed at this time?


Homework Equations


TotalForcex=ma
Delta x=(1/2)(delta v)(delta t)
Delta v=a(delta t)


The Attempt at a Solution


1. Total force of x is 250N, so 250N=ma=250N/40= acceleration = 6.25m/s^2
2. (deltax)=(1/2)(6.25)(5.35)^2 = 89.4
3. Vf=(6.25)(5.35)=33.4 m/s


Does this look alright?

The astronaut and asteroid are both in space. Don't forget Newton's 3rd Law.

For every action there is a reaction. The 250N force on the asteroid is also a 250N force on the Astronaut in the opposite direction. You need to also take into account the acceleration on the astronaut. The question asks for their distance apart after 5s and their relative speed. (One is going in the opposite direction to the other.)

As to method, you need to figure first the distance apart after the Force stops, then there is the constant velocity over 5s.
 
  • #3
LowlyPion said:
The astronaut and asteroid are both in space. Don't forget Newton's 3rd Law.

For every action there is a reaction. The 250N force on the asteroid is also a 250N force on the Astronaut in the opposite direction. You need to also take into account the acceleration on the astronaut. The question asks for their distance apart after 5s and their relative speed. (One is going in the opposite direction to the other.)

As to method, you need to figure first the distance apart after the Force stops, then there is the constant velocity over 5s.


So one question before I get going, my professor had said that the total time would be 5.35 and that the wording of the problem is just a bit confusing. Does this sound right? Thanks much for the help!
 
  • #4
soul_observer said:
So one question before I get going, my professor had said that the total time would be 5.35 and that the wording of the problem is just a bit confusing. Does this sound right? Thanks much for the help!

It could be seen as ambiguous. The intention is clear though that there is the initial .35s acceleration phase and the 5 second constant velocity phase.
 
  • #5
Okay, so I began by analyzing the total forces of both the astronaut and the asteroid. The total forces on the astronaut = -250N = ma so I then found the acceleration of by dividing -250 by mass 60kg to obtain a=-4.17. Doing the same thing for the asteroid I used a postive 250/mass 40kg and obtained an a=6.25. Then I went into obtaining the velocity of both objects so that I could proceed to find the x value. I used the equation deltaV=a(t). I used t=0.35s And for the astronaut I obtained v= -1.46 and asteroid v=2.19. Then I proceed to find x using the equation x=(1/2)(deltaV). The answers I obtained for x were pretty small. So I'm guessing I'm on the wrong track... Do you know what I should or shouldn't do?
 
  • #6
soul_observer said:
Okay, so I began by analyzing the total forces of both the astronaut and the asteroid. The total forces on the astronaut = -250N = ma so I then found the acceleration of by dividing -250 by mass 60kg to obtain a=-4.17. Doing the same thing for the asteroid I used a postive 250/mass 40kg and obtained an a=6.25. Then I went into obtaining the velocity of both objects so that I could proceed to find the x value. I used the equation deltaV=a(t). I used t=0.35s And for the astronaut I obtained v= -1.46 and asteroid v=2.19. Then I proceed to find x using the equation x=(1/2)(deltaV). The answers I obtained for x were pretty small. So I'm guessing I'm on the wrong track... Do you know what I should or shouldn't do?

The difference in acceleration between the two is 6.25 and -(-4.17) which is a total of 10.42.

With x = 1/2*a*t2 is .63m

With V2 = 2*a*x that means that the relative V = 3.65 m/s times 5 sec ... plus x ...
 
  • #7
Ah ha! I see, okay okay; let's see if I can do this! Thank you so so so much!
 
  • #8
Okay, got it. Thank you Very much; I would have been lost without you!
 

1. How is the velocity of the astronaut pushing an asteroid calculated?

The velocity of the astronaut pushing an asteroid can be calculated using the formula v = d/t, where v is the velocity in meters per second, d is the distance covered in meters, and t is the time taken in seconds.

2. What factors affect the velocity of the astronaut pushing an asteroid?

The velocity of the astronaut pushing an asteroid can be affected by a number of factors, such as the mass and size of the asteroid, the strength and technique of the astronaut, and any external forces acting on the asteroid, such as gravity or friction.

3. How is the distance covered by the astronaut pushing an asteroid determined?

The distance covered by the astronaut pushing an asteroid can be determined by measuring the distance between the starting point and the endpoint of the asteroid's movement. This can be done using a measuring tool, such as a ruler or tape measure.

4. Is it possible for the astronaut to calculate their velocity and distance covered while pushing the asteroid?

Yes, it is possible for the astronaut to calculate their velocity and distance covered while pushing the asteroid. This can be done by recording the time taken to push the asteroid and measuring the distance covered, then using the formula v = d/t to calculate the velocity.

5. Can the velocity and distance covered by the astronaut pushing an asteroid be affected by other factors?

Yes, the velocity and distance covered by the astronaut pushing an asteroid can be affected by other factors, such as the surface on which the asteroid is being pushed, the gravitational pull of nearby objects, and the physical condition of the astronaut. These factors should be taken into consideration when conducting experiments or calculations.

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