Physics Solution: Car Stopping Distance

The problem involves calculating the stopping distance of a car at 72 kmph, given that it stops in 5 m from 36 kmph under the same braking conditions (no skid).

Stopping Distance Physics Principles

Under constant deceleration (from non-skidding brakes), the stopping distance ($s$) is related to initial velocity ($u$) by the kinematic equation $v^2 = u^2 + 2as$. With final velocity $v=0$, we get $0 = u^2 + 2as$, which simplifies to $s = -u^2/(2a)$.

This shows that stopping distance ($s$) is directly proportional to the square of the initial velocity ($u^2$):

$ s \propto u^2 $

Speed Conversion

Convert speeds from kmph to m/s using the conversion factor $1 \text{ kmph} = \frac{5}{18} \text{ m/s}$:

• Initial speed $u_1 = 36 \text{ kmph} = 36 \times \frac{5}{18} = 10 \text{ m/s}$.
• New speed $u_2 = 72 \text{ kmph} = 72 \times \frac{5}{18} = 20 \text{ m/s}$.

Stopping Distance Calculation

Let $s_1$ be the stopping distance at $u_1$ and $s_2$ be the stopping distance at $u_2$. The ratio of stopping distances is equal to the ratio of the squares of the initial velocities:

$ \frac{s_2}{s_1} = \left( \frac{u_2}{u_1} \right)^2 $

Substitute the given values:

• $s_1 = 5$ m
• $u_1 = 10$ m/s
• $u_2 = 20$ m/s

$ \frac{s_2}{5 \text{ m}} = \left( \frac{20 \text{ m/s}}{10 \text{ m/s}} \right)^2 $

$ \frac{s_2}{5 \text{ m}} = (2)^2 = 4 $

Calculate $s_2$:

$ s_2 = 4 \times 5 \text{ m} = 20 \text{ m} $

Conclusion

The car will stop in 20 meters when going at 72 kmph.