Respuesta :
(a) 5.35 s, 84.65 s
The distance covered by the rocket during the first part of the motion (accelerated motion) is
[tex]d_1 = \frac{1}{2}at_1 ^2[/tex] (1)
where
a = +35 ft/s^2 is the acceleration
While the distance covered during the second part of the motion (uniform motion) is
[tex]d_2 = v t_2[/tex]
where v is the constant speed reached after the first part of the motion, which is given by
[tex]v=at_1[/tex]
So we can rewrite the equation as
[tex]d_2 = a t_1 t_2[/tex] (2)
We also know:
[tex]d= d_1 + d_2 = 16350[/tex] (3) is the total distance
[tex]t= t_1 +t_2 = 90[/tex] (4) is the total time
We can rewrite (3) as
[tex]\frac{1}{2}at_1^2 + a t_1 t_2 = 16350[/tex] (5)
And (4) as
[tex]t_2 = 90-t_1[/tex]
Substituting into (5), we get
[tex]\frac{1}{2}at_1^2 + a t_1 (90-t_1) = 16350\\0.5 at_1^2 + 90at_1 - at_1^2 = 16350\\-0.5at_1^2 +90at_1 = 16350[/tex]
Substituting a=+35 ft/s^2, we have
[tex]-17.5 t_1^2 +3150 t_1 - 16350 = 0[/tex]
which has two solutions:
[tex]t_1 = 174.65 s[/tex] --> discarded, since the total time cannot be greater than 90 s
[tex]t_1 = 5.35 s[/tex] --> this is the correct solution
And therefore [tex]t_2[/tex] is
[tex]t_2 = 90-t_1 = 90 -5.35 s=84.65 s[/tex]
(b) 187.25 ft/s
The velocity v is the velocity at the end of the accelerated motion of the rocket, so after
[tex]t_1 = 5.35 s[/tex]
The acceleration is
a = +35 ft/s^2
Therefore, the velocity after the first part is:
[tex]v=at_1 = (+35 ft/s^2)(5.35 s)=187.25 ft/s[/tex]
(c) 16,585 ft
At the 15750 ft mark, the velocity of the sled is
[tex]v=187.25 ft/s[/tex]
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:
[tex]v_f ^ 2 -v^2 = 2ad_3[/tex]
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,
[tex]d_3 = \frac{v_f^2 -v^2}{2a}=\frac{0-(187.25 ft/s)^2}{2(-21 ft/s^2)}=835 ft[/tex]
So, the final position of the sled will be
[tex]x_3 = 15750 ft + 835 ft =16,585 ft[/tex]
(d) 93.92 s
The duration of the first part of the motion is
[tex]t_1 = 5.35 s[/tex]
The distance covered during this part is
[tex]d_1 = \frac{1}{2}at_1^2=\frac{1}{2}(+35 ft/s^2)(5.35 s)^2=501 ft[/tex]
In the second part (uniform motion), the sled continues with constant speed until the 15750 ft mark, so the distance covered is
[tex]d_2 = 15750- 835 =14915[/tex]
And so the duration of the second part is
[tex]t_2 =\frac{d_2}{v}=\frac{14915}{187.25}=79.65 s[/tex]
While the duration of the third part (decelerated motion) is
[tex]a=\frac{v_f-v}{t_3}\\t_3 = \frac{v_f-f}{a}=\frac{0-(187.25 ft/s)}{-21 ft/s^2}=8.92 s[/tex]
So the total duration is
t = 5.35 s+79.65 s+8.92 s=93.92 s
