A particle had a speed of 18 m/s at a certain time, and 2.5 s later its speed was 30 m/s in the opposite direction. The average acceleration of the particle in the duration is :

Options

A. ` 19.2 m//s^(2) ` in the direction of initial velocity
B. ` 19.2 m//s^(2) ` in the direction opposite to the initial velocity
C. ` 4.8 m//s^(2) ` in the direction of initial velocity
D. ` 4.8 m//s^(2) ` in the direction opposite to the initial velocity

Explanation

To solve the problem of finding the average acceleration of the particle, we can follow these steps:

### Step 1: Identify the initial and final velocities
- The initial speed of the particle (V_initial) is given as 18 m/s.
- The final speed of the particle (V_final) is given as 30 m/s in the opposite direction.

### Step 2: Assign directions
- We will consider the initial direction (where the particle is moving at 18 m/s) as positive.
- Therefore, the final speed of 30 m/s in the opposite direction will be negative. Thus:
- V_initial = +18 m/s
- V_final = -30 m/s

### Step 3: Calculate the change in velocity
- The change in velocity (ΔV) can be calculated as:
\[ \Delta V = V_f - V_i = (-30) - (18) = -30 - 18 = -48 \text{ m/s} \]

### Step 4: Determine the time interval
- The time interval (Δt) during which this change occurs is given as 2.5 seconds.

### Step 5: Calculate the average acceleration
- The average acceleration (a_avg) can be calculated using the formula:
\[ a_{avg} = \frac{\Delta V}{\Delta t} = \frac{-48 \text{ m/s}}{2.5 \text{ s}} = -19.2 \text{ m/s}^2 \]

### Step 6: Interpret the result
- The negative sign indicates that the average acceleration is in the opposite direction to the initial velocity.

### Final Answer
- The average acceleration of the particle is **-19.2 m/s²**.

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