QJEEM18Ph96b8: JEE Main 2018 Physics - Class 11 JEEM18Ph96b8 Solution: An automobile, travelling at km/h, can be stopped at a distance …

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.