A driver travelling at speed `36 kmh^(-1)` sees the light trun red at the intersection. If his reaction time is 0.6s, and then the car can deaccelerate at `4 ms^(-2)`. Find the stopping distance of the car.

Explanation

To find the stopping distance of the car, we will break down the problem into several steps:

### Step 1: Convert the speed from km/h to m/s
The speed of the car is given as \( 36 \, \text{km/h} \). We need to convert this speed into meters per second (m/s).

\[ \text{Speed in m/s} = 36 \times \frac{5}{18} = 10 \, \text{m/s} \]

### Step 2: Calculate the distance traveled during the reaction time
The driver's reaction time is given as \( 0.6 \, \text{s} \). During this time, the car continues to travel at its initial speed of \( 10 \, \text{m/s} \).

\[ \text{Distance during reaction time} = \text{Speed} \times \text{Time} = 10 \, \text{m/s} \times 0.6 \, \text{s} = 6 \, \text{m} \]

### Step 3: Determine the deceleration phase
After the reaction time, the car begins to decelerate at \( 4 \, \text{m/s}^2 \). The initial velocity at this point is still \( 10 \, \text{m/s} \), and the final velocity when the car stops is \( 0 \, \text{m/s} \).

### Step 4: Use the kinematic equation to find the stopping distance
We will use the kinematic equation:

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

Where:
- \( v \) is the final velocity (0 m/s),
- \( u \) is the initial velocity (10 m/s),
- \( a \) is the acceleration (-4 m/s², since it is deceleration),
- \( s \) is the distance we want to find.

Rearranging the equation to solve for \( s \):

\[ 0 = (10)^2 + 2 \times (-4) \times s \]

This simplifies to:

\[ 0 = 100 - 8s \]

Solving for \( s \):

\[ 8s = 100 \implies s = \frac{100}{8} = 12.5 \, \text{m} \]

### Step 5: Calculate the total stopping distance
The total stopping distance is the sum of the distance traveled during the reaction time and the distance traveled while decelerating.

\[ \text{Total stopping distance} = \text{Distance during reaction time} + \text{Distance while decelerating} \]
\[ \text{Total stopping distance} = 6 \, \text{m} + 12.5 \, \text{m} = 18.5 \, \text{m} \]

### Final Answer
The stopping distance of the car is \( 18.5 \, \text{m} \).

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