A car travelling with a speed 125 KMPH along a straight line comes to rest after travelling a distance 245 m.The time taken by the car to come to rest,in second is

Options

A. 11
B. 12
C. 16
D. 14

Answer

To solve the problem step by step, we will follow the principles of motion in a straight line.

### Step 1: Convert the speed from km/h to m/s
The initial speed of the car is given as 125 km/h. To convert this to meters per second (m/s), we use the conversion factor:

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

Calculating this:

\[ \text{Initial speed} (U) = 125 \times \frac{5}{18} = 34.72 \, \text{m/s} \]

### Step 2: Identify the final speed
Since the car comes to rest, the final speed (V) is:

\[ V = 0 \, \text{m/s} \]

### Step 3: Use the first equation of motion
We can use the first equation of motion to find the acceleration (a):

\[ V = U + at \]

Substituting the known values:

\[ 0 = 34.72 + a \cdot t \]

Rearranging gives:

\[ a \cdot t = -34.72 \quad \text{(1)} \]

### Step 4: Use the third equation of motion
Next, we will use the third equation of motion to relate distance (s), initial speed (U), time (t), and acceleration (a):

\[ s = Ut + \frac{1}{2} a t^2 \]

Substituting the known values:

\[ 245 = 34.72 \cdot t + \frac{1}{2} a t^2 \quad \text{(2)} \]

### Step 5: Substitute acceleration from equation (1) into equation (2)
From equation (1), we can express acceleration (a) in terms of time (t):

\[ a = -\frac{34.72}{t} \]

Substituting this into equation (2):

\[ 245 = 34.72 \cdot t + \frac{1}{2} \left(-\frac{34.72}{t}\right) t^2 \]

This simplifies to:

\[ 245 = 34.72t - \frac{34.72}{2} t \]

\[ 245 = 34.72t - 17.36t \]

\[ 245 = 17.36t \]

### Step 6: Solve for time (t)
Now, we can solve for t:

\[ t = \frac{245}{17.36} \approx 14.11 \, \text{seconds} \]

### Final Answer
Thus, the time taken by the car to come to rest is approximately:

\[ t \approx 14 \, \text{seconds} \]

---