A vehicle is moving on a rough horizontal road with velocity v. The stopping distance will be directly proportional to

Correct Option: v²

To find the relationship between the stopping distance of a vehicle and its initial velocity on a rough horizontal road, we can apply the principles of mechanics.

Let:
- $v$ be the initial velocity of the vehicle,
- $d$ be the stopping distance,
- $m$ be the mass of the vehicle,
- $\mu$ be the coefficient of friction between the tires and the road, and
- $g$ be the acceleration due to gravity.

When the brakes are applied, the frictional force ($f$) acting on the vehicle opposes its motion and brings it to a stop. The frictional force on a horizontal road is given by:

$f=\mu mg$

According to Newton's second law of motion, this force produces a constant retardation (deceleration) $a$ given by:

$a=-\frac{f}{m}=-\mu g$

We can now use the third equation of motion to relate the initial velocity, final velocity, acceleration, and distance:

${v_f}^2=u^2+2ad$

Here, the initial velocity $u=v$, and the final velocity $v_f=0$ (since the vehicle comes to rest). Substituting these values along with the acceleration:

$0^2=v^2+2(-\mu g)d$

Simplifying the equation:

$v^2=2\mu gd$

Solving for the stopping distance $d$:

$d=\frac{v^2}{2\mu g}$

Since the coefficient of friction $\mu$ and the acceleration due to gravity $g$ are constants for a given vehicle and road surface, the expression reveals that:

$d\propto v^2$

Therefore, the stopping distance is directly proportional to the square of the velocity ($v^2$).