1) 9.18 s
In the first part of the motion, the rocket accelerates at a rate of
[tex]a_1=13.5 m/s^2[/tex]
For a time period of
[tex]t_1=3.50 s[/tex]
So we can calculate the velocity of the rocket after this time period by using the SUVAT equation:
[tex]v_1=u+a_1t_1[/tex]
where u = 0 is the initial velocity of the rocket. Substituting a1 and t1,
[tex]v_1=(13.5)(3.50)=47.3 m/s[/tex]
In the second part of the motion, the rocket decelerates with a constant acceleration of
[tex]a_2 = -5.15 m/s^2[/tex]
Until it comes to a stop, to reach a final velocity of
[tex]v_2 = 0[/tex]
So we can use again the same equation
[tex]v_2 = v_1 + a_2 t_2[/tex]
where [tex]v_1 = 47.3 m/s[/tex]. Solving for t2, we find after how much time the rocket comes to a stop:
[tex]t_2 = -\frac{v_1}{a_2}=-\frac{47.3}{5.15}=9.18 s[/tex]
2) 299.9 m
We have to calculate the distance travelled by the rocket in each part of the motion.
The distance travelled in the first part is given by:
[tex]d_1 = ut_1 + \frac{1}{2}a_1 t_1^2[/tex]
Using the numbers found in part a),
[tex]d_1 = 0 + \frac{1}{2}(13.5) (3.50)^2=82.7 m[/tex]
The distance travelled in the second part of the motion is
[tex]d_2= v_1 t_2 + \frac{1}{2}a_2 t_2^2[/tex]
Using the numbers found in part a),
[tex]d_2 = (47.3)(9.18) + \frac{1}{2}(-5.15) (9.18)^2=217.2 m[/tex]
So, the total distance travelled by the rocket is
d = 82.7 m + 217.2 m = 299.9 m
