At a certain time a particle had a speed of 18 m/s in the positive `x` direction, and 2.4 s later its speed was 30 m/s in the opposite direction. What is the average acceleration of the particle during this 2.4 s interval ?

Answer

To find the average acceleration of the particle during the 2.4 seconds interval, we can follow these steps:

### Step 1: Identify the initial and final velocities
- The initial velocity \( v_i \) is given as \( 18 \, \text{m/s} \) in the positive \( x \) direction.
- The final velocity \( v_f \) is given as \( 30 \, \text{m/s} \) in the opposite direction, which we will represent as \( -30 \, \text{m/s} \).

### Step 2: Write the formula for average acceleration
The formula for average acceleration \( a \) is given by:
\[ a = \frac{v_f - v_i}{t} \]
where:
- \( v_f \) is the final velocity,
- \( v_i \) is the initial velocity,
- \( t \) is the time interval.

### Step 3: Substitute the values into the formula
Substituting the values we have:
\[ a = \frac{-30 \, \text{m/s} - 18 \, \text{m/s}}{2.4 \, \text{s}} \]

### Step 4: Simplify the equation
Calculating the numerator:
\[ -30 \, \text{m/s} - 18 \, \text{m/s} = -48 \, \text{m/s} \]
Now substituting this back into the equation for acceleration:
\[ a = \frac{-48 \, \text{m/s}}{2.4 \, \text{s}} \]

### Step 5: Calculate the average acceleration
Now, performing the division:
\[ a = -20 \, \text{m/s}^2 \]

### Conclusion
The average acceleration of the particle during the 2.4 seconds interval is \( -20 \, \text{m/s}^2 \).

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