A car moving with a speed of 40 km/h can be stopped by applying brakes after atleast 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance? (a) 8 m (b) 2 m (c) 4 m (d) 6 m

Answer: (a). For the same retardation, s = u²/2a ⇒ s ∝ u². Speed doubles (40 to 80 km/h), so distance becomes 2² = 4 times. s' = 4 × 2 = 8 m So the answer is (a) 8 m.

A car moving with a speed of 40 km/h can be stopped by applying… | NEET 1998 (AIPMT)

Q41NEET 1998 · AIPMTEasy

Equations of Kinematics

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

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Answer: (a)

For the same retardation, $s = \frac{u^2}{2a} \Rightarrow s \propto u^2$ .
Speed doubles (40 to 80 km/h), so distance becomes $2^2 = 4$ times.
$s' = 4 \times 2 = 8\text{ m}$
So the answer is (a) $8\text{ m}$ .

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