Explanation:
Let
[tex]x_1[/tex] = distance traveled while accelerating
[tex]x_2[/tex] = distance traveled while decelerating
The distance traveled while accelerating is given by
[tex]x_1 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2[/tex]
[tex]\:\:\:\:\:= \frac{1}{2}(2.5\:\text{m/s}^2)(30\:\text{s})^2[/tex]
[tex]\:\:\:\:\:= 1125\:\text{m}[/tex]
We need the velocity of the rocket after 30 seconds and we can calculate it as follows:
[tex]v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}[/tex]
This will be the initial velocity when start calculating for the distance it traveled while decelerating.
[tex]v^2 = v_0^2 + 2ax_2[/tex]
[tex]0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2[/tex]
Solving for [tex]x_2,[/tex] we get
[tex]x_2 = \dfrac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:\:= 4327\:\text{m}[/tex]
Therefore, the total distance x is
[tex]x = x_1 + x_2 = 1125\:\text{m} + 4327\:\text{m}[/tex]
[tex]\:\:\:\:= 5452\:\text{m}[/tex]
