A person travelling at 43.2 km/h applies the brake giving a deceleration of `6.0 m/ss^2` to his scooter. How far will it travel before stopping?

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Convert the initial velocity from km/h to m/s
The initial velocity \( u \) is given as \( 43.2 \, \text{km/h} \). We need to convert this to meters per second (m/s).

\[ u = 43.2 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = \frac{43200 \, \text{m}}{3600 \, \text{s}} = 12 \, \text{m/s} \]

### Step 2: Identify the deceleration
The deceleration \( a \) is given as \( 6.0 \, \text{m/s}^2 \). Since it is a deceleration, we will consider it as negative in our calculations.

\[ a = -6.0 \, \text{m/s}^2 \]

### Step 3: Use the third equation of motion
We will use the third equation of motion, which relates initial velocity, final velocity, acceleration, and distance traveled. The equation is:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) is the final velocity (0 m/s, since the scooter stops)
- \( u \) is the initial velocity (12 m/s)
- \( a \) is the acceleration (-6 m/s²)
- \( s \) is the distance traveled before stopping

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:

\[ 0 = (12)^2 + 2(-6)s \]

This simplifies to:

\[ 0 = 144 - 12s \]

### Step 5: Solve for \( s \)
Rearranging the equation to solve for \( s \):

\[ 12s = 144 \]

\[ s = \frac{144}{12} = 12 \, \text{m} \]

### Conclusion
The distance the scooter will travel before stopping is \( 12 \, \text{m} \).

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