A car is going with a linear momentum $p$ . When brakes are applied, it comes to a stop in a distance $s$ . If the same car were going with a linear momentum $2 p$ and the brakes are applied, it comes to a stop in a distance of (assume that the brake force is same in the two cases)

2 s

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s

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4 s

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\frac{s}{4}

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Solution

The correct option is C $4 s$
$v^{2} = 2 a s_{1}$
$4 v^{2} = 2 a s_{2}$
a is same for both since mass is same and applied force is same.
$8 a s_{1} = 2 a s_{2}$
$\Rightarrow s_{2} = 4 s_{1}$

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A car is going with a linear momentum p. When brakes are applied, it comes to a stop in a distance s. If the same car were going with a linear momentum 2p and the brakes are applied, it comes to a stop in a distance of (assume that the brake force is same in the two cases)

The correct option is C 4sv2=2as14v2=2as2a is same for both since mass is same and applied force is same.8as1=2as2 ⇒s2=4s1

A car is going with a linear momentum $p$ . When brakes are applied, it comes to a stop in a distance $s$ . If the same car were going with a linear momentum $2 p$ and the brakes are applied, it comes to a stop in a distance of (assume that the brake force is same in the two cases)

The correct option is C $4 s$
$v^{2} = 2 a s_{1}$
$4 v^{2} = 2 a s_{2}$
a is same for both since mass is same and applied force is same.
$8 a s_{1} = 2 a s_{2}$
$\Rightarrow s_{2} = 4 s_{1}$