Question

At a time when mining asteroids has become feasible, astronauts have connected a line between their 3740-kg space tug and a 5690-kg asteroid. They pull on the asteroid with a force of 518 N. Initially the tug and the asteroid are at rest, 432 m apart. How much time does it take for the ship and the asteroid to meet?

Answer 1

Answer:

61.4 s

Explanation:

The distance d₁ traveled by the asteroid:

$d_1=\frac{1}{2}a_1t^2$

The distance d₂ traveled by the space ship:

$d_2=\frac{1}{2}a_2t^2$

The total distance d:

$d=d_1+d_2=\frac{1}{2}(a_1+a_2)t^2$

Solving for time t:

$t=\sqrt{\frac{2d}{a_1+a_2}}=\sqrt{\frac{2d}{F(\frac{1}{m_1}+\frac{1}{m_2})}}=\sqrt{\frac{2dm_1m_2}{F(m_1+m_2)}}$

Answer 2

Answer:

The answer is 61.35 s

Explanation:

We start by writing the distance equations:

d = xtug + xasteroid (eq. 1)

xtug = votug*t + (atug*t^2)/2; if votug = 0, we have:

xtug = (atug*t^2)/2

xasteroid = voasteroid*t +(aasteroid*t^2)/2; if voasteroid = 0, we have:

xasteroid = (aasteroid*t^2)/2

replacing in equation 1:

d = (atug*t^2)/2 + (aasteroid*t^2)/2 = (atug + aasteroid)*t^2/2

Clearing t:

t = ((2*d)/(atug + aasteroid))^1/2

applying Newton's second law we have:

F = m*a; F = force, m = mass, a=acceleration

a= F/m

t = ((2*d)/(F*((1/mtug) +(1/masteroid))))^1/2 = ((2*432)/(518*((1/3740) + (1/5690))))^1/2 = 61.35 s