A vehicle of mass 1000 kg is moving with a velocity of `15 ms^(-1)`.It is brought to rest by applying brakes and locking the wheels. If the slidding friction between the tyres and the road is 6000 N, then the distance moved by the vehicle before coming to rest is

Options

A. `37.5 m`
B. `18.75 m`
C. `75 m`
D. `15 m`

Explanation

To solve the problem step by step, we will use the concepts of physics related to motion, friction, and the equations of motion.

### Step 1: Identify the given values
- Mass of the vehicle (m) = 1000 kg
- Initial velocity (u) = 15 m/s
- Frictional force (F_friction) = 6000 N

### Step 2: Calculate the retardation (deceleration)
Using Newton's second law, the retardation (a) can be calculated using the formula:
\[ F = m \cdot a \]
Rearranging this gives:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{6000 \, \text{N}}{1000 \, \text{kg}} = 6 \, \text{m/s}^2 \]
Since this is a deceleration, we will take it as negative:
\[ a = -6 \, \text{m/s}^2 \]

### Step 3: Use the third equation of motion to find the distance
The third equation of motion relates initial velocity (u), final velocity (v), acceleration (a), and distance (s):
\[ v^2 = u^2 + 2as \]
Here, the final velocity (v) when the vehicle comes to rest is 0 m/s. Substituting the known values:
\[ 0 = (15)^2 + 2 \cdot (-6) \cdot s \]
This simplifies to:
\[ 0 = 225 - 12s \]
Rearranging gives:
\[ 12s = 225 \]
Thus:
\[ s = \frac{225}{12} = 18.75 \, \text{m} \]

### Conclusion
The distance moved by the vehicle before coming to rest is **18.75 meters**.

---