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 step-by-step, we will follow these steps:

### Step 1: Convert the speed from km/h to m/s
The initial speed of the car is given as \( u = 144 \, \text{km/h} \). We need to convert this speed into meters per second (m/s).

\[ u = 144 \, \text{km/h} \times \frac{5}{18} = 40 \, \text{m/s} \]

### Step 2: Identify the final velocity and distance
The car comes to a stop, so the final velocity \( v = 0 \, \text{m/s} \). The distance \( s \) over which the car stops is given as \( s = 200 \, \text{m} \).

### Step 3: Use the third equation of motion
We can use the third equation of motion, which relates initial velocity, final velocity, acceleration, and distance:

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

Substituting the known values:

\[ 0 = (40)^2 + 2a(200) \]

### Step 4: Solve for acceleration \( a \)
Rearranging the equation gives:

\[ 0 = 1600 + 400a \]

This simplifies to:

\[ 400a = -1600 \]

So,

\[ a = -4 \, \text{m/s}^2 \]

### Step 5: Use the first equation of motion to find time \( t \)
Now we will use the first equation of motion to find the time taken to stop:

\[ v = u + at \]

Substituting the known values:

\[ 0 = 40 + (-4)t \]

Rearranging gives:

\[ 4t = 40 \]

So,

\[ t = \frac{40}{4} = 10 \, \text{s} \]

### Final Answer
The time taken for the car to stop is \( t = 10 \, \text{seconds} \).

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