Question

URGENT!!!!!!! PLEASE HELP WITH THIS PHYSICS PROBLEM

Answer 1

Explanation:

Let

$x_1$ = distance traveled while accelerating

$x_2$ = distance traveled while decelerating

The distance traveled while accelerating is given by

$x_1 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2$

$\:\:\:\:\:= \frac{1}{2}(2.5\:\text{m/s}^2)(30\:\text{s})^2$

$\:\:\:\:\:= 1125\:\text{m}$

We need the velocity of the rocket after 30 seconds and we can calculate it as follows:

$v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}$

This will be the initial velocity when start calculating for the distance it traveled while decelerating.

$v^2 = v_0^2 + 2ax_2$

$0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2$

Solving for $x_2,$ we get

$x_2 = \dfrac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}$

$\:\:\:\:\:= 4327\:\text{m}$

Therefore, the total distance x is

$x = x_1 + x_2 = 1125\:\text{m} + 4327\:\text{m}$

$\:\:\:\:= 5452\:\text{m}$