Answer:
The astronaut travels 49.0 meters before coming to a stop with respect to the spacecraft.
Step-by-step explanation:
We can use the conservation of momentum to solve this problem:
The initial momentum of the astronaut is:
p1 = m * v1 = 63 kg * 7.0 m/s = 441 kg·m/s to the right
When the astronaut comes to a stop with respect to the spacecraft, the final momentum is:
p2 = m * v2 = 63 kg * 0 m/s = 0
The change in momentum is:
Δp = p2 - p1 = -441 kg·m/s
The time taken for the astronaut to come to a stop is t = 14.0 s.
Therefore, the average force exerted by the jet pack can be found using the formula:
F = Δp / t
F = (-441 kg·m/s) / (14.0 s) = -31.5 N
The negative sign indicates that the force is in the opposite direction to the initial motion of the astronaut, as expected.
The distance traveled by the astronaut before coming to a stop can be found using the formula:
d = (1/2) * a * t^2
where a is the acceleration, which is equal to the force divided by the mass:
a = F / m = (-31.5 N) / (63 kg) = -0.5 m/s^2
Substituting the values we get:
d = (1/2) * (-0.5 m/s^2) * (14.0 s)^2 = 49.0 m
Therefore, the astronaut travels 49.0 meters before coming to a stop with respect to the spacecraft.