At a time when mining asteroids has become feasible, astronauts have connected a line between their 3460-kg space tug and a 6430-kg asteroid. They pull on the asteroid with a force of 366 N. Initially the tug and the asteroid are at rest, 493 m apart. How much time does it take for the ship and the asteroid to meet?

77.8s

Explanation:

Let d distance between the asteroid and space tug

So;d=Xtug+Xspace

Xtug=VtT+0.5atT^2

Xspace=VsT+0.5asT^2

Since Vt=Vs=0 initial velocity

Then

d=0.5(atT^2+asT^2)

T^2( at+as)=2d

T=√(2d/at+as)

But force F = mass M*acceleration a

Hence at=Ft/mt ,as=Fs/ms

But note Ft=F=Fs since the Same force acts on it

Hence T=√( 2d/F(1/mt+1/Ms))

T=√(2*493/366(1/3460+1/6430)

T=√(986/0.1627)=√(6060.195)=77.8s

the inner core.

good luck: )

klicker