Kinematics An elevator is traveling at a constant speed of 2.6 m/s. At time t = 0 s, an object is dropped from the elevator shaft. After 2.7 s has passed, the distance between the object and the elevator is _____ m

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Kinematics

An elevator is traveling at a constant speed of 2.6 m/s. At time t = 0 s, an object is dropped from the elevator shaft. After 2.7 s has passed, the distance between the object and the elevator is _____ m

**Problem Statement:**

An elevator is traveling at a constant speed of 2.6 m/s. At time \( t = 0 \) seconds, an object is dropped from the elevator shaft. After 2.7 seconds have passed, the distance between the object and the elevator is ______ meters.

**Explanation:**

To solve this problem, consider the following:

1. **Elevator Movement:**
   - The elevator continues to move downward at 2.6 m/s. Therefore, in 2.7 seconds, the elevator travels:
     \[
     \text{Distance by elevator} = \text{speed} \times \text{time} = 2.6 \, \text{m/s} \times 2.7 \, \text{s} = 7.02 \, \text{m}
     \]

2. **Object Movement:**
   - The object is in free fall, so its distance fallen can be calculated using the formula for distance under constant acceleration due to gravity:
     \[
     \text{Distance fallen by object} = \frac{1}{2} g t^2
     \]
     where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
     \[
     \text{Distance fallen by object} = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (2.7 \, \text{s})^2 = 35.595 \, \text{m}
     \]

3. **Distance Between Object and Elevator:**
   - The distance between the object and the elevator after 2.7 seconds is the difference between the distance fallen by the object and the distance traveled by the elevator:
     \[
     \text{Distance between object and elevator} = 35.595 \, \text{m} - 7.02 \, \text{m} = 28.575 \, \text{m}
     \]

Therefore, after 2.7 seconds, the distance between the object and the elevator is approximately 28.575 meters.
Transcribed Image Text:**Problem Statement:** An elevator is traveling at a constant speed of 2.6 m/s. At time \( t = 0 \) seconds, an object is dropped from the elevator shaft. After 2.7 seconds have passed, the distance between the object and the elevator is ______ meters. **Explanation:** To solve this problem, consider the following: 1. **Elevator Movement:** - The elevator continues to move downward at 2.6 m/s. Therefore, in 2.7 seconds, the elevator travels: \[ \text{Distance by elevator} = \text{speed} \times \text{time} = 2.6 \, \text{m/s} \times 2.7 \, \text{s} = 7.02 \, \text{m} \] 2. **Object Movement:** - The object is in free fall, so its distance fallen can be calculated using the formula for distance under constant acceleration due to gravity: \[ \text{Distance fallen by object} = \frac{1}{2} g t^2 \] where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. \[ \text{Distance fallen by object} = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (2.7 \, \text{s})^2 = 35.595 \, \text{m} \] 3. **Distance Between Object and Elevator:** - The distance between the object and the elevator after 2.7 seconds is the difference between the distance fallen by the object and the distance traveled by the elevator: \[ \text{Distance between object and elevator} = 35.595 \, \text{m} - 7.02 \, \text{m} = 28.575 \, \text{m} \] Therefore, after 2.7 seconds, the distance between the object and the elevator is approximately 28.575 meters.
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