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Answer :
To solve this problem, we need to use the equations of motion, which are applicable under constant acceleration (or deceleration, as is the case here).
Given:
- Initial velocity [tex]v_i = 40.0 \, \text{m/s}[/tex]
- Final velocity [tex]v_f = 0 \, \text{m/s}[/tex] (since the airplane comes to a stop)
- Deceleration [tex]a = -1.50 \, \text{m/s}^2[/tex] (it's negative because it is deceleration)
We have two objectives:
(a) To find the length of the runway needed:
We can use the equation:
[tex]v_f^2 = v_i^2 + 2a s[/tex]
where:
- [tex]s[/tex] is the distance (length of the runway)
Rearranging the formula to solve for [tex]s[/tex], we get:
[tex]s = \frac{v_f^2 - v_i^2}{2a}[/tex]
Substitute the values:
[tex]s = \frac{0^2 - (40.0)^2}{2(-1.50)}[/tex]
[tex]s = \frac{-1600}{-3.00}[/tex]
[tex]s = 533.33 \, \text{m}[/tex]
So, the airplane needs approximately 533.33 meters of runway.
(b) For how long will it move on the runway?
We can use the following equation:
[tex]v_f = v_i + at[/tex]
where:
- [tex]t[/tex] is the time
Rearrange to find [tex]t[/tex]:
[tex]t = \frac{v_f - v_i}{a}[/tex]
Substitute the values:
[tex]t = \frac{0 - 40.0}{-1.50}[/tex]
[tex]t = \frac{-40.0}{-1.50}[/tex]
[tex]t = 26.67 \, \text{s}[/tex]
Therefore, the airplane will be moving on the runway for approximately 26.67 seconds.
In summary, the airplane will require about 533.33 meters of runway length and will decelerate to a stop in about 26.67 seconds.
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