A ball hits a wall horizontally at 6.0 m `s^(-1)`. It rebounds horzontally at 4.4 m `s^(-1)`. The ball is in contact with the wall for 0.040 s. What is the acceleration of the ball ?

Options

A. 260 units
B. 264 units
C. - 40 units
D. None of these

Explanation

To find the acceleration of the ball after it hits the wall, we can follow these steps:

### Step 1: Identify the initial and final velocities
- The initial velocity (V1) of the ball when it hits the wall is **6.0 m/s** (towards the wall).
- The final velocity (V2) of the ball after it rebounds is **4.4 m/s** (away from the wall).

### Step 2: Assign signs to the velocities
- Since the ball is moving towards the wall initially, we can assign it a negative sign: **V1 = -6.0 m/s**.
- After rebounding, the ball is moving away from the wall, so we assign it a positive sign: **V2 = 4.4 m/s**.

### Step 3: Calculate the change in velocity
- The change in velocity (ΔV) can be calculated using the formula:
\[ \Delta V = V2 - V1 \]
- Substituting the values:
\[ \Delta V = 4.4 - (-6.0) = 4.4 + 6.0 = 10.4 \, \text{m/s} \]

### Step 4: Use the time of contact
- The time (T) during which the ball is in contact with the wall is given as **0.040 seconds**.

### Step 5: Calculate the acceleration
- The acceleration (a) can be calculated using the formula:
\[ a = \frac{\Delta V}{T} \]
- Substituting the values:
\[ a = \frac{10.4 \, \text{m/s}}{0.040 \, \text{s}} = 260 \, \text{m/s}^2 \]

### Step 6: Determine the direction of acceleration
- Since the ball is decelerating while hitting the wall (the change in velocity is positive but the initial velocity was negative), we can conclude that the acceleration is in the opposite direction of the initial velocity. Thus, we represent it as:
\[ a = -260 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the ball is **-260 m/s²**.

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