Consider a car moving along a straight horizontal road with a speed of 72 km/hr. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is ( 10 m/s² )

The correct option is 40 m.

To find the shortest distance in which the car can be stopped, we can analyze the motion using the equations of kinematics and the concept of friction.

Step 1: Convert the initial velocity of the car into SI units (m/s)
The initial velocity ($u$) is given as 72 km/hr. To convert km/hr to m/s, we multiply by $\frac{5}{18}$:
$u=72\times \frac{5}{18}\text{m/s}=20\text{m/s}$

Step 2: Determine the maximum deceleration of the car
The stopping force is provided by the static friction between the tyres and the road. The maximum force of static friction ($f_s$) is:
$f_s=\mu N=\mu mg$
where $\mu$ is the coefficient of static friction, $m$ is the mass of the car, and $g$ is the acceleration due to gravity.

According to Newton's second law of motion, the maximum retarding acceleration (deceleration, $a$) is:
$a=\frac{f_s}{m}$
Substituting the given values:
$a=0.5\times 10{\text{m/s}}^2=5{\text{m/s}}^2$

Step 3: Calculate the stopping distance
Since the car is coming to a stop, its final velocity ($v$) is 0. Using the third equation of motion:
$v^2=u^2-2as$
where $s$ is the stopping distance. Substituting $v=0$:
$0=u^2-2as$
$s=\frac{u^2}{2a}$

Substitute the values of $u$ and $a$ into the equation:
$s=\frac{{20}^2}{2\times 5}=\frac{400}{10}=40\text{m}$

Therefore, the shortest distance in which the car can be stopped is 40 m.