An automobile traveling at 40 km/h can be stopped within a distance of 40 meters by using brakes. If the same automobile is traveling at 80 km/h, we need to determine the minimum stopping distance in meters, assuming no skidding occurs.

First Case

• Initial speed, $ u_1 = 40 \, \text{km/h} $


• Final speed, $ v_1 = 0 \, \text{m/s} $


• Stopping distance, $ s_1 = 40 \, \text{m} $

Using the equation:

$ v^2 - u^2 = 2as $

For the first case:

$ 0^2 - 40^2 = 2a \times 40 $

$ -1600 = 80a $

$ a = -20 \, \text{m/s}^2 $

Second Case

• Initial speed, $ u_2 = 80 \, \text{km/h} $


• Final speed, $ v_2 = 0 \, \text{m/s} $

Similarly, using the same equation:

$ 0^2 - 80^2 = 2a s_2 $

$ -6400 = 2a s_2 $

Since $ a = -20 \, \text{m/s}^2 $, divide both sides of the equation for the second case by the first case:

$ \frac{s_2}{40} = \frac{80^2}{40^2} $

$ s_2 = \frac{80 \times 80}{40} $

$ s_2 = 160 \, \text{m} $

Thus, the minimum stopping distance when the automobile is traveling at 80 km/h is 160 meters.