Question

An automobile travelling with a speed of 60 km/h , can brake to stop within a distance of 20m . If the car is going twice as fast . i.e, 120km/h , the stopping distance will be [AIEEE 2004] ​

Answer 1

Ratio of Stopping distance:

$= \dfrac{ {u1}^{2} }{2a} \div \dfrac{ {u2}^{2} }{2a}$

$\implies \dfrac{20m}{x} = \dfrac{60 \times 60}{2a} \div \dfrac{120 \times 120}{2a}$

$\implies \: \dfrac{20}{x} = \dfrac{3600}{2a} \times \dfrac{2a}{120 \times 120}$

$\implies \: \dfrac{20}{x} = \dfrac{1}{4}$

$\red{\bold{\boxed{ \implies \: x = 80m}}}$

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PS: Here we don't need to look at the units as we're comparing by taking the ratios

Answer 2

Explanation:- The breaking retardation will remain same and assumed to be constant , let it be a .

• From third equation of motion
• V² = u²+2as

case ___(1)

• 0² = (60×5/18)² -2as₁
• => s₁ = (60 × 5/18)²/2as₂

case ______(2)

• 0 = (120 × 5/18)² -2as₂
• => s₂ =( 120 × 5/18)²/2a
• ∴ s₁/s₂ = 1/4
• => s₂ = 4s₁ = 4* 20 = 80 m Answer

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