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URGENT!!!!!!!

PLEASE HELP WITH THIS PHYSICS PROBLEM

URGENT!!!!!!! PLEASE HELP WITH THIS PHYSICS PROBLEM-example-1

1 Answer

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Step-by-step 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 + (1)/(2)at^2 = (1)/(2)at^2


\:\:\:\:\:= (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 = \frac{(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}

User VinothRaja
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