If a man increases his speed by 2 , m/s,his k | vedclass

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(C) Let the initial speed of the man be $v$ and his mass be $m$.Initial kinetic energy is $E = \frac{1}{2}mv^2$ ... $(i)$When the speed increases by $

Vedclass · Accepted Answer

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

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(C) Let the initial speed of the man be $v$ and his mass be $m$.Initial kinetic energy is $E = \frac{1}{2}mv^2$ ... $(i)$When the speed increases by $2 \, m/s$,the new speed becomes $(v + 2) \, m/s$.The new kinetic energy is $2E = \frac{1}{2}m(v + 2)^2$ ... (ii)Dividing equation (ii) by equation $(i)$:$\frac{2E}{E} = \frac{\frac{1}{2}m(v + 2)^2}{\frac{1}{2}mv^2}$$2 = \frac{(v + 2)^2}{v^2}$Taking the square root on both sides:$\sqrt{2} = \frac{v + 2}{v}$$v\sqrt{2} = v + 2$$v(\sqrt{2} - 1) = 2$$v = \frac{2}{\sqrt{2} - 1}$Rationalizing the denominator:$v = \frac{2(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2\sqrt{2} + 2}{2 - 1} = 2 + 2\sqrt{2} \, m/s$.

If a man increases his speed by $2 \, m/s$,his kinetic energy is doubled. The original speed of the man is:

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