fisica1-youn-e-freedman-exercicios-resolvidos

8.107: a) For t < 0 the rocket is at rest. For ≤ t ≤ 8 .40 is valid, and
v( t) = ( 2400 m s) ln( 1 ( 1−
( t /120 s)
)).
At t = 90 s, this speed is 3.33 km/s, and this is also
the speed for t > 90 s.
0 90 s, Eq. ( )
b) The acceleration is zero for t < 0 and t > 90 s. For 0 ≤ t ≤ 90 s, Eq. (8.39) gives,
with dm
2
20 m/s
= −m0
120 s, a =
dt
( 1−( t /120 s)
) .
c) The maximum acceleration occurs at the latest time of firing, t = 90 s, at which time
2
20 m/s
2
the acceleration is, from the result of part (a), ( ) = 80 m/s , and so the astronaut is
1−
90 /120
subject to a force of 6.0 kN, about eight times her weight on earth.