A pitcher delivers a fast ball with a velocity of 43 m/s to the south. The batter hits the ball and gives it a velocity of 51 m/s to the north. What was the average acceleration of the ball during the 1.0 ms when it was in contact with the bat ?

Options

A. `1.5` m/s
B. `34 m//s`
C. `7.0 m//s`
D. `49 m//s`

Answer

To solve the problem of finding the average acceleration of the ball during the 1.0 ms it was in contact with the bat, we will follow these steps:

### Step 1: Identify the given values
- Initial velocity of the ball (v_initial) = 43 m/s (to the south)
- Final velocity of the ball (v_final) = 51 m/s (to the north)
- Time of contact (Δt) = 1.0 ms = 1.0 x 10^-3 s

### Step 2: Define the direction of the velocities
- We will consider south as negative and north as positive for our calculations:
- v_initial = -43 m/s (south)
- v_final = +51 m/s (north)

### Step 3: Calculate the change in velocity (Δv)
The change in velocity is given by the formula:
\[ \Delta v = v_{final} - v_{initial} \]
Substituting the values:
\[ \Delta v = 51 \, \text{m/s} - (-43 \, \text{m/s}) = 51 \, \text{m/s} + 43 \, \text{m/s} = 94 \, \text{m/s} \]

### Step 4: Calculate the average acceleration (a)
The average acceleration is defined as the change in velocity divided by the time taken:
\[ a = \frac{\Delta v}{\Delta t} \]
Substituting the values:
\[ a = \frac{94 \, \text{m/s}}{1.0 \times 10^{-3} \, \text{s}} = 94,000 \, \text{m/s}^2 \]

### Step 5: Determine the direction of the acceleration
Since the final velocity is in the north direction (positive) and the initial velocity is in the south direction (negative), the acceleration is also directed towards the north.

### Final Answer
The average acceleration of the ball during the 1.0 ms when it was in contact with the bat is:
\[ 94,000 \, \text{m/s}^2 \, \text{(to the north)} \]

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