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
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?
1 answer
1 answer