If the speed of a vehicle increases by $2 m/s$ , its kinetic energy is doubled, then original speed of the vehicle is

\sqrt{2} + 1 m/s

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2 (\sqrt{2} + 1) m/s

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2 (\sqrt{2} - 1) m/s

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\sqrt{2} (\sqrt{2} + 1) m/s

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Solution

The correct option is B $2 (\sqrt{2} + 1) m/s$
$K_{f} = 2 K_{i}$
$\frac{1}{2} m (v + 2)^{2} = (\frac{1}{2} m v^{2}) (2)$
$v + 2 = \sqrt{2} v$
$v = \frac{2}{\sqrt{2} - 1} = 2 (\sqrt{2} + 1) m/s$

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If the speed of a vehicle increases by 2 m/s, its kinetic energy is doubled, then original speed of the vehicle is

The correct option is B 2(√2+1)m/sKf=2Ki 12m(v+2)2=(12mv2)(2) v+2=√2v v=2√2−1=2(√2+1)m/s

If the speed of a vehicle increases by $2 m/s$ , its kinetic energy is doubled, then original speed of the vehicle is

The correct option is B $2 (\sqrt{2} + 1) m/s$
$K_{f} = 2 K_{i}$
$\frac{1}{2} m (v + 2)^{2} = (\frac{1}{2} m v^{2}) (2)$
$v + 2 = \sqrt{2} v$
$v = \frac{2}{\sqrt{2} - 1} = 2 (\sqrt{2} + 1) m/s$