An elevator is moving upward with a speed of 11 m/s. Three seconds later, the elevator is still moving upward, but its speed has been reduced to 5.0 m/s. What is the average acceleration of the elevator during the 3.0 s interval ?

Options

A. The particle remains at rest after `t=3.0` s.
B. The particle no longer accelerates after `=3.0` s.
C. The particle can be found at positions `x lt 0` m only when `t lt 0` s.
D. None of the above is correct .

Answer

To solve the problem of finding the average acceleration of the elevator during the 3.0 s interval, we can follow these steps:

### Step 1: Identify the given values
- Initial velocity (u) = 11 m/s (upward)
- Final velocity (v) = 5 m/s (upward)
- Time interval (t) = 3 s

### Step 2: Write the formula for average acceleration
The average acceleration (a) can be calculated using the formula:
\[ a = \frac{\Delta v}{\Delta t} \]
where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time.

### Step 3: Calculate the change in velocity
The change in velocity (\(\Delta v\)) is given by:
\[ \Delta v = v - u \]
Substituting the values:
\[ \Delta v = 5 \, \text{m/s} - 11 \, \text{m/s} = -6 \, \text{m/s} \]

### Step 4: Substitute the values into the acceleration formula
Now, substitute \(\Delta v\) and \(\Delta t\) into the acceleration formula:
\[ a = \frac{-6 \, \text{m/s}}{3 \, \text{s}} = -2 \, \text{m/s}^2 \]

### Step 5: Interpret the result
The negative sign indicates that the acceleration is directed downward, opposite to the direction of the initial velocity (upward). Therefore, the average acceleration of the elevator is:
\[ a = -2 \, \text{m/s}^2 \quad \text{(downward)} \]

### Final Answer
The average acceleration of the elevator during the 3.0 s interval is \(2 \, \text{m/s}^2\) downward.

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