(a) 5.35 s, 84.65 s
The distance covered by the rocket during the first part of the motion (accelerated motion) is
(1)
where
a = +35 ft/s^2 is the acceleration
While the distance covered during the second part of the motion (uniform motion) is
where v is the constant speed reached after the first part of the motion, which is given by
So we can rewrite the equation as
(2)
We also know:
(3) is the total distance
(4) is the total time
We can rewrite (3) as
(5)
And (4) as
Substituting into (5), we get
Substituting a=+35 ft/s^2, we have
which has two solutions:
--> discarded, since the total time cannot be greater than 90 s
--> this is the correct solution
And therefore is
(b) 187.25 ft/s
The velocity v is the velocity at the end of the accelerated motion of the rocket, so after
The acceleration is
a = +35 ft/s^2
Therefore, the velocity after the first part is:
(c) 16,585 ft
At the 15750 ft mark, the velocity of the sled is
Then, it starts decelerating with acceleration
a = -21 ft/s^2
So, its distance covered during this part of the motion will be given by:
where
vf = 0 is the final speed
v = 187.25 ft/s is the initial speed
a = -21 ft/s^2 is the acceleration
d3 is the distance covered during this part
Solving for d3,
So, the final position of the sled will be
(d) 93.92 s
The duration of the first part of the motion is
The distance covered during this part is
In the second part (uniform motion), the sled continues with constant speed until the 15750 ft mark, so the distance covered is
And so the duration of the second part is
While the duration of the third part (decelerated motion) is
So the total duration is
t = 5.35 s+79.65 s+8.92 s=93.92 s