Step-by-step explanation:
Let
= distance traveled while accelerating
= distance traveled while decelerating
The distance traveled while accelerating is given by
![x_1 = v_0t + (1)/(2)at^2 = (1)/(2)at^2](https://img.qammunity.org/2022/formulas/physics/college/9kbvkownay34lew0692oyvackqpmkmeg0i.png)
![\:\:\:\:\:= (1)/(2)(2.5\:\text{m/s}^2)(30\:\text{s})^2](https://img.qammunity.org/2022/formulas/physics/college/xrca1ops6ze5b440t01yoz02pk6crrsrne.png)
![\:\:\:\:\:= 1125\:\text{m}](https://img.qammunity.org/2022/formulas/physics/college/hbbu15qf41kb7i7mk1war8tgrul8lr4s8l.png)
We need the velocity of the rocket after 30 seconds and we can calculate it as follows:
![v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}](https://img.qammunity.org/2022/formulas/physics/college/nwdlgksadiw0ofg5k5phdww00aez9tfwvb.png)
This will be the initial velocity when start calculating for the distance it traveled while decelerating.
![v^2 = v_0^2 + 2ax_2](https://img.qammunity.org/2022/formulas/physics/college/ghw1tm6jrtca96e30kl105ony4hdwp30hk.png)
![0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2](https://img.qammunity.org/2022/formulas/physics/college/hssxwe8kgvarbd7vn0ubh8zrn05fgl49je.png)
Solving for
we get
![x_2 = \frac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}](https://img.qammunity.org/2022/formulas/physics/college/pr4md9krgnbe6kzltequu5n57cltsfht2c.png)
![\:\:\:\:\:= 4327\:\text{m}](https://img.qammunity.org/2022/formulas/physics/college/l4uzbo7ni7czv2qs12djryn2gd34nuwvw6.png)
Therefore, the total distance x is
![x = x_1 + x_2 = 1125\:\text{m} + 4327\:\text{m}](https://img.qammunity.org/2022/formulas/physics/college/3dhav4zdr6mftwzaw50gimwv26tehgkjcp.png)
![\:\:\:\:= 5452\:\text{m}](https://img.qammunity.org/2022/formulas/physics/college/7f5s17zwjaxmjo5gi3lg8lgb7ep095ccir.png)