An automobile, travelling at 40km/h, can be stopped at a distance of 40m by applying brakes. If the same automobile is travelling at 80km/h, the minimum stopping distance, in meters, is (assuming no skidding)
A. 75m
B. 160m
C. 100m
D. 150m

Hint: This is a simple kinematic problem which can be easily solved using Newton's laws of motion formulae in one dimension. The Newton’s laws of motion formulae in one dimension give relations between final velocity, initial velocity, acceleration, distance and time. Remember, it’s the same automobile travelling at different speeds.

Complete step-by-step answer:
Here we have two cases, one where the automobile is travelling at initial velocity of 40km/h and the next case where the automobile is travelling at 80km/h.

For the first case, we have ${{u}_{1}}=40km/h,{{v}_{1}}=0km/h,{{d}_{1}}=40m,a=?$

Where ${{u}_{1}},{{v}_{1}},{{d}_{1}}$ and $a$ denote the initial velocity, final velocity, distance after the application of brakes until the automobile comes to a halt and the acceleration of the automobile respectively.

Just one Newton’s law of motion is required \[{{v}^{2}}-{{u}^{2}}=2ad\]

Using this by substituting the first case variables, $v_{1}^{2}-u_{1}^{2}=2a{{d}_{1}}$

Before using the same value of ${{u}_{1}}$ and ${{v}_{1}},$ we should first convert them to m/s from km/h as the solution for distance is in meters.

Conversion from km/h to m/s in $xkm/h=\dfrac{5x}{18}m/s$

Using this, ${{u}_{1}}=40km/h=\dfrac{40\times 5}{18}m/s=\dfrac{200}{18}m/s$
Therefore, ${{0}^{2}}-{{(\dfrac{200}{18})}^{2}}=2a(40)$

Hence, the braking acceleration is $a=\dfrac{-500}{18}m/{{s}^{2}}$

Moving on to the second case, where the initial velocity of the automobile is 80 km/h.
${{u}_{2}}=80km/h=\dfrac{80\times
5}{18}m/s=\dfrac{400}{18}m/s,{{v}_{2}}=0km/h=0m/s,a=\dfrac{-500}{18\times 18}m/{{s}^{2}}$
Using the same formula of \[{{v}^{2}}-{{u}^{2}}=2ad\]
${{0}^{2}}-{{(\dfrac{400}{18})}^{2}}=2(\dfrac{-500}{{{18}^{2}}}){{d}_{2}}$
$\dfrac{160000}{1000}m={{d}_{2}}$

Therefore, the minimum stopping distance in the second case is 160 m.

Note: We should remember that the same automobile is travelling at different speeds in both the cases hence, the other variables except the braking acceleration will remain constant.
Braking negative acceleration or deceleration, hence, don’t worry about the negative sign.