Consider, a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between the tyre and the road is 0.5, the shortest distance in which the car can be stopped is (Take `g=10 //s^(2)`)

Options

A. 30m
B. 40m
C. 72m
D. 20m

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the speed from km/h to m/s
The initial speed \( U \) of the car is given as 72 km/h. We need to convert this speed into meters per second (m/s).

\[ U = 72 \text{ km/h} = \frac{72 \times 1000 \text{ m}}{3600 \text{ s}} = 20 \text{ m/s} \]

### Step 2: Calculate the maximum frictional force
The maximum static frictional force \( f \) that can act on the car is given by:

\[ f = \mu \cdot R \]

Where:
- \( \mu \) is the coefficient of static friction (given as 0.5)
- \( R \) is the normal reaction force, which is equal to the weight of the car \( mg \) (where \( m \) is the mass of the car and \( g \) is the acceleration due to gravity, taken as \( 10 \, \text{m/s}^2 \)).

Thus, we can express the frictional force as:

\[ f = \mu \cdot mg \]

### Step 3: Relate frictional force to acceleration
According to Newton's second law, the frictional force is also equal to the mass of the car multiplied by its acceleration \( a \):

\[ f = ma \]

Setting the two expressions for frictional force equal gives:

\[ \mu mg = ma \]

We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

\[ \mu g = a \]

### Step 4: Calculate the acceleration
Substituting the values of \( \mu \) and \( g \):

\[ a = 0.5 \cdot 10 = 5 \, \text{m/s}^2 \]

Since this is a deceleration (the car is stopping), we will consider it as negative:

\[ a = -5 \, \text{m/s}^2 \]

### Step 5: Use the equation of motion to find the stopping distance
We will use the third equation of motion:

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

Where:
- \( v \) is the final velocity (0 m/s, since the car stops)
- \( u \) is the initial velocity (20 m/s)
- \( a \) is the acceleration (-5 m/s²)
- \( s \) is the distance we want to find

Substituting the known values:

\[ 0 = (20)^2 + 2(-5)s \]

This simplifies to:

\[ 0 = 400 - 10s \]

### Step 6: Solve for \( s \)
Rearranging gives:

\[ 10s = 400 \]

\[ s = \frac{400}{10} = 40 \, \text{m} \]

Thus, the shortest distance in which the car can be stopped is:

\[ \boxed{40 \, \text{m}} \]

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