A man moves on his motorbike with `54 km//h` and then takes a U-turn and containues to move with same velocity The time of U-turn is `10 s`. Find the magnitude of average acceleration during U-turn.

Options

A. 0
B. `3ms^(-2)`
C. `1.5sqrt(2)ms^(-2)`
D. none of these

Explanation

To find the magnitude of average acceleration during the U-turn, we can follow these steps:

### Step 1: Convert the speed from km/h to m/s
The speed of the motorbike is given as 54 km/h. We need to convert this to meters per second (m/s) using the conversion factor \(1 \text{ km/h} = \frac{5}{18} \text{ m/s}\).

\[ \text{Speed in m/s} = 54 \times \frac{5}{18} = 15 \text{ m/s} \]

### Step 2: Identify initial and final velocities
When the man takes a U-turn, he initially moves in one direction and then turns around to move in the opposite direction.

- **Initial velocity (v_initial)**: When moving in the positive direction, we can represent this as \(+15 \text{ m/s}\).
- **Final velocity (v_final)**: After the U-turn, he moves in the opposite direction, which can be represented as \(-15 \text{ m/s}\).

### Step 3: Calculate the change in velocity
The change in velocity (\(\Delta v\)) can be calculated as:

\[ \Delta v = v_{final} - v_{initial} = (-15 \text{ m/s}) - (15 \text{ m/s}) = -30 \text{ m/s} \]

### Step 4: Calculate average acceleration
Average acceleration (\(a_{avg}\)) is defined as the change in velocity divided by the time taken for that change. The time taken for the U-turn is given as 10 seconds.

\[ a_{avg} = \frac{\Delta v}{\Delta t} = \frac{-30 \text{ m/s}}{10 \text{ s}} = -3 \text{ m/s}^2 \]

### Step 5: Find the magnitude of average acceleration
The magnitude of average acceleration is the absolute value of the average acceleration:

\[ \text{Magnitude of } a_{avg} = | -3 \text{ m/s}^2 | = 3 \text{ m/s}^2 \]

### Final Answer
The magnitude of average acceleration during the U-turn is \(3 \text{ m/s}^2\).

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