A car moving along a straight road with speed of `144 km h^(-1)` is brought to a stop within a distance of 200 m. How long does it take for the car to stop ?

Options

A. 5 s
B. 10 s
C. 15 s
D. 20 s

Explanation

To solve the problem of how long it takes for a car moving at a speed of 144 km/h to stop within a distance of 200 m, we can follow these steps:

### Step 1: Convert the speed from km/h to m/s
The initial speed \( u \) is given as 144 km/h. To convert this speed to meters per second (m/s), we use the conversion factor:
\[ 1 \text{ km/h} = \frac{5}{18} \text{ m/s} \]
Thus,
\[ u = 144 \times \frac{5}{18} = 40 \text{ m/s} \]

### Step 2: Identify the final velocity
When the car comes to a stop, its final velocity \( v \) is:
\[ v = 0 \text{ m/s} \]

### Step 3: Use the kinematic equation to find acceleration
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity = 0 m/s
- \( u \) = initial velocity = 40 m/s
- \( a \) = acceleration (which will be negative since it's deceleration)
- \( s \) = distance = 200 m

Substituting the known values into the equation:
\[ 0 = (40)^2 + 2a(200) \]
\[ 0 = 1600 + 400a \]
Rearranging gives:
\[ 400a = -1600 \]
\[ a = -\frac{1600}{400} = -4 \text{ m/s}^2 \]

### Step 4: Use the acceleration to find the time taken to stop
Now we can use the equation:
\[ v = u + at \]
Substituting the known values:
\[ 0 = 40 + (-4)t \]
Rearranging gives:
\[ 4t = 40 \]
\[ t = \frac{40}{4} = 10 \text{ seconds} \]

### Conclusion
The time taken for the car to stop is:
\[ \boxed{10 \text{ seconds}} \]