Question

85. (IIT-JEE 2003) A car moving with
JEE 2003) A car moving with a speed of 50 km/h
can be stopped by brakes after at least 6 m. If the same
car is moving at a speed of 100 km/h, the minimum
stopping distance is
(a).12 m (b) 18 m
(c) 24 m (d) 6 m​

Answer 1

Answer:

First Case:

Initial Speed of the car: 50 km/hr

Final Speed of the car: 0 km/hr

Stopping Distance = 6m

Acceleration = ?

Using the third equation of motion we get that,

⇒ v² - u² = 2as

⇒ 50² - 0² = 2a × 6

⇒ 2500 = 12a

⇒ a = 2500/12 m/s²    ... ( Eqn 1 )

Second Case:

Initial Speed = 100 km/hr

Final Speed = 0 km/hr

Acceleration = 2500/12 m/s²

Stopping distance = ?

Using the third equation of motion again we get that,

⇒ 100² - 0² = 2 × 2500/12 × s

⇒ 10000 = 5000/12 × s

⇒ s = 10000 × 12 / 5000 = 24 m

Hence the minimum stopping distance is 24 m.

Answer 2

$\huge\bf{Answer}$

According to the First Case:-

• 50 km/hr is the Speed of the car.
• 0 km/hr is the Final Speed of the car.
• 6m is the Stopping Distance.

Let us find the what will be the Acceleration !!

v² - u² = 2as

50² - 0² = 2a × 6

2500 = 12a

a = 2500/12 m/s²  This is Equation - (1)

According to the Second Case:-

• 100 km/hr is the Speed of the car.
• 0 km/hr final Speed of the car.

In this Acceleration is given

• 2500/12 m/s² is the Acceleration

Let us find the Stopping distance

100² - 0² = 2 × 2500/12 × s

10000 = 5000/12 × s

s = 10000 × 12 / 5000 = 24 m

Therefore ,24 m is the minimum stopping distance.