At a time when mining asteroids has become feasible, astronauts have connected a line between their 3360-kg space tug and a 6300-kg asteroid. Using their ship's engine, they pull on the asteroid with a force of 490 N. Initially the tug and the asteroid are at rest, 430 m apart. How much time does it take for the ship and the asteroid to meet?

F = m a

Ft = tug force = - Fa asteroid force

3360 vt + 6300 va = 0 momentum remains 0

490 = 3360 at
-490 = 6300 aa

at = .146 m/s^2
aa = -0.778 m/s^2

vt = 0 + .146 t
va = 0 - .778 t

position t = .5 (.146)t^2
430 - position a = .5 (-.778)t^2

they are headed toward each other
so when does the sum = 430 meters?

.5 t^2 (.146+.778) = 430
solve for t

To find the time it takes for the ship and the asteroid to meet, we need to use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a), or F = ma.

First, let's calculate the acceleration of the system. The total mass of the system is the sum of the mass of the space tug (3360 kg) and the mass of the asteroid (6300 kg), which is 9660 kg.

F = ma
490 N = 9660 kg * a

Now, we can solve for the acceleration (a):

a = 490 N / 9660 kg
a ≈ 0.0507 m/s²

Next, we can use the equation of motion, s = ut + 0.5at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.

In this case, the initial velocity (u) is 0 m/s since both the tug and the asteroid are initially at rest.

We want to find the time it takes for the ship and the asteroid to meet when they are 430 m apart. So, we can rearrange the equation to solve for time (t):

s = ut + 0.5at²
430 m = 0 * t + 0.5 * 0.0507 m/s² * t²

Simplifying the equation further:

430 m = 0.02535 m/s² * t²

Now, we can solve for time (t):

t² = 430 m / (0.02535 m/s²)
t² ≈ 16994.116

Taking the square root of both sides:

t ≈ √(16994.116)
t ≈ 130.44 seconds

So, it takes approximately 130.44 seconds for the ship and the asteroid to meet.