Question

If the velocity of a car is increased by 20% then the minimum distance in which the car can be stopped increases by how much percentage?

Answer 1

Minimum distance (S) is given as
S = u² / (2a)

For a car with acceleration “a”
S ∝ u²

S₂/S₁ = (u₂/u₁)²
S₂/S₁ - 1 = (u₂/u₁)² - 1
ΔS / S₁ = (u₂/u₁)² - 1

ΔS = S₁ × [(u₂/u₁)² - 1]
= 100 × [(120 / 100)² - 1]
= 100 × (144/100 - 1)
= 44

Minimum distance increases by 44%

Answer 2

Answer:

44%

Explanation:

S = u² / (2a)

S ∝ u²

S₂/S₁ = (u₂/u₁)²

S₂/S₁ - 1 = (u₂/u₁)² - 1

ΔS / S₁ = (u₂/u₁)² - 1

ΔS = S₁ × [(u₂/u₁)² - 1]

= 100 × [(120 / 100)² - 1]

= 100 × (144/100 - 1)

= 44