A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is

To find the minimum stopping distance of the car when its speed is doubled, we can use the equations of motion under uniform retardation (deceleration).

The third equation of motion is given by:
$v^2=u^2+2as$
where:
- $v$ is the final velocity of the car (which is 0 when the car stops),
- $u$ is the initial velocity of the car,
- $a$ is the acceleration (which is negative deceleration, $-a$), and
- $s$ is the stopping distance.

Substituting $v=0$ and acceleration as $-a$ into the equation:
$0=u^2-2as$
This simplifies to:
$u^2=2as$
Or:
$s=\frac{u^2}{2a}$

Assuming the braking force, and therefore the deceleration ($a$), remains constant in both cases, the stopping distance $s$ is directly proportional to the square of the initial speed $u$:
$s\propto u^2$

This relationship can be expressed as a ratio for two different cases:
$\frac{s_2}{s_1}={\left(\frac{u_2}{u_1}\right)}^2$

Given:
- Initial speed in the first case, $u_1=50\text{km/hr}$
- Stopping distance in the first case, $s_1=6\text{m}$
- Initial speed in the second case, $u_2=100\text{km/hr}$

Now, substitute these values into the ratio:
$\frac{s_2}{6}={\left(\frac{100}{50}\right)}^2$
$\frac{s_2}{6}={\left(2\right)}^2$
$\frac{s_2}{6}=4$
$s_2=4\times 6$
$s_2=24\text{m}$

Therefore, the minimum stopping distance is 24m.