Question

A car moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance?

• A. 4 m
• B. 6 m
• C. 8 m
• D. 2 m

Hint:

From motion's third equation, we get \(v^2 - u^2 = 2as\).

Answer 1

The Correct Option is C

For the 1^st case, u = 40 km/h = \(40 \times \frac{5}{18} \text{ m/s}\)

From motion's third equation, we get

v² - u² = 2as

v² =  u² + 2as

\(0 = \left(40 \times \frac{5}{18}\right)^2 + 2a(2)\)

\(a = -\frac{1}{4}\left(\frac{100}{9}\right)^2 - 1\)

For the 2nd case - 

u = 80 km/h 

v² - u² = 2as

v² =  u² + 2as

\(0 = \left(80 \times \frac{5}{18}\right)^2 + 2as\)

On solving the equation, we get s=8m.

Hence, option (C) is the correct answer.

Answer 2

Two conditions given in the question are:

1. The speed of the car = 40 km/h, after applying brakes, the car can be stopped after 2m.
2. The speed of the car = 80 km/h.

Here we have only the car's speed and distance, and time is not given to us. So, we use the third equation of motion to calculate the required minimum stopping distance.

v² =  u² + 2as

• where u = initial speed of the car
• v = final speed of the car
• a = acceleration of the car
• s = distance travelled by car.

We know that the car's final speed will become zero (0). Substituting v = 0 km/h and the values of the first condition in the above equation, we get,

\(0=(40km/h\times\frac{1000m}{1km}\times\frac{1h}{3600sec})^{2}−2a(2m)\)

\(a=\frac{123.45m^{2}/s^{2}}{4m}\)

a = 30.86 m/s²

We will use the acceleration value with the given values of the second condition to get the minimum stopping distance when the speed of the car is 80 km/h.

Therefore, from the third equation of motion, we get,

\(0=(80km/h\times\frac{1000m}{1km}×\frac{1h}{3600sec})^{2}−2(30.86m/s^{2}){s}\)

\(s=\frac{493.83m^{2}/s^{2}}{61.72m/s^{2}}\)

a = 8m

Hence, the minimum stopping distance is 8m, and therefore option (C) is the correct answer.

Answer 3

Stopping distance, s = \(u^2 \over 2a\)

\({s_2 \over s_1} = {u_2^2 \over u_1^2}\)

\({s_2} = {[{u_2 \over u_1}]^2}s_1\)

s₂ = (80/40)²2

s = 8m

A car moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. If the same car is moving with a speed of 80 km/h, the minimum stopping distance is 8 m.