An airplane starts from rest and accelerates at 10.8 m/s?. What is its speed at the end of a 500 m-long runway? 37.0 m/s 103.9 m/s O 65.7 m/s 93.0 m/s O 186 m/s

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Chapter1: Units, Trigonometry. And Vectors
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**Understanding Acceleration and Final Speed**

*Problem Statement:*

An airplane starts from rest and accelerates at \(10.8 \, \text{m/s}^2\). What is its speed at the end of a \(500 \, \text{m}\)-long runway?

*Options:*
- ○ \(37.0 \, \text{m/s}\)
- ○ \(103.9 \, \text{m/s}\)
- ○ \(65.7 \, \text{m/s}\)
- ○ \(93.0 \, \text{m/s}\)
- ○ \(186 \, \text{m/s}\)

---

### Solution Approach

To find the final speed (\(v\)) of the airplane at the end of the runway, we can use the kinematic equation:

\[ v^2 = u^2 + 2as \]

where:
- \( u \) is the initial velocity (which is \(0 \, \text{m/s}\) since the plane starts from rest),
- \( a \) is the acceleration (\(10.8 \, \text{m/s}^2\)),
- \( s \) is the displacement (length of the runway, \(500 \, \text{m}\)).

Plugging in the values:

\[ v^2 = 0 + 2 \cdot 10.8 \, \text{m/s}^2 \cdot 500 \, \text{m} \]
\[ v^2 = 10800 \, \text{m}^2/\text{s}^2 \]
\[ v = \sqrt{10800} \, \text{m/s} \]
\[ v \approx 103.9 \, \text{m/s} \]

Thus, the correct option is \(103.9 \, \text{m/s}\).

---

**Learning Points:**
- This problem helps understand how to apply kinematic equations to calculate the final velocity of an object when given initial velocity, acceleration, and distance.
- This exercise emphasizes the importance of understanding motion under constant acceleration and how to manipulate the kinematic equations accordingly.
Transcribed Image Text:**Understanding Acceleration and Final Speed** *Problem Statement:* An airplane starts from rest and accelerates at \(10.8 \, \text{m/s}^2\). What is its speed at the end of a \(500 \, \text{m}\)-long runway? *Options:* - ○ \(37.0 \, \text{m/s}\) - ○ \(103.9 \, \text{m/s}\) - ○ \(65.7 \, \text{m/s}\) - ○ \(93.0 \, \text{m/s}\) - ○ \(186 \, \text{m/s}\) --- ### Solution Approach To find the final speed (\(v\)) of the airplane at the end of the runway, we can use the kinematic equation: \[ v^2 = u^2 + 2as \] where: - \( u \) is the initial velocity (which is \(0 \, \text{m/s}\) since the plane starts from rest), - \( a \) is the acceleration (\(10.8 \, \text{m/s}^2\)), - \( s \) is the displacement (length of the runway, \(500 \, \text{m}\)). Plugging in the values: \[ v^2 = 0 + 2 \cdot 10.8 \, \text{m/s}^2 \cdot 500 \, \text{m} \] \[ v^2 = 10800 \, \text{m}^2/\text{s}^2 \] \[ v = \sqrt{10800} \, \text{m/s} \] \[ v \approx 103.9 \, \text{m/s} \] Thus, the correct option is \(103.9 \, \text{m/s}\). --- **Learning Points:** - This problem helps understand how to apply kinematic equations to calculate the final velocity of an object when given initial velocity, acceleration, and distance. - This exercise emphasizes the importance of understanding motion under constant acceleration and how to manipulate the kinematic equations accordingly.
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