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(D) Let the initial velocity be $v$. Initial kinetic energy $E = \frac{1}{2}mv^2$. When the velocity is increased by $1 \,m/s$, the new velocity becom

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

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(D) Let the initial velocity be $v$. Initial kinetic energy $E = \frac{1}{2}mv^2$. When the velocity is increased by $1 \,m/s$, the new velocity becomes $v' = v + 1$. The new kinetic energy $E' = \frac{1}{2}m(v+1)^2$. According to the problem, the kinetic energy is doubled, so $E' = 2E$. Substituting the expressions: $\frac{1}{2}m(v+1)^2 = 2 	imes (\frac{1}{2}mv^2)$. Simplifying the equation: $(v+1)^2 = 2v^2$. Expanding the left side: $v^2 + 2v + 1 = 2v^2$. Rearranging the terms to form a quadratic equation: $v^2 - 2v - 1 = 0$. Using the quadratic formula $v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $v = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$. Since velocity must be positive, we take $v = (1 + \sqrt{2}) \,m/s$.

The kinetic energy of a car is doubled when its velocity is increased by $1 \,m/s$. Then the initial velocity of the car is

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