Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is [μk = 0.5] (a) 1000 m (b) 800 m (c) 400 m (d) 100 m

Answer: (a). Retardation a = μk g = 0.5 × 10 = 5 m/s² (taking g = 10 m/s²). s = v²/2a = ((100)²)/(10) s = 1000 m, so the answer is (a).

Consider a car moving on a straight road with a speed of 100 m/s. The… | JEE Main 2005 (AIEEE)

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Friction Basics

Consider a car moving on a straight road with a speed of $100\text{ m/s}$ . The distance at which car can be stopped is $[\mu_k = 0.5]$

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Answer: (a)

Retardation $a = \mu_k g = 0.5 \times 10 = 5\text{ m/s}^2$ (taking $g = 10\text{ m/s}^2$ ).
$s = \dfrac{v^2}{2a} = \dfrac{(100)^2}{10}$
$s = 1000$ m, so the answer is (a).