A man who is running has half the kinetic ene | vedclass

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(D) Let the mass of the man be $M$ and his initial speed be $v$. Let the mass of the boy be $m = M/2$ and his speed be $u$.Given that the kinetic ener

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

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(D) Let the mass of the man be $M$ and his initial speed be $v$. Let the mass of the boy be $m = M/2$ and his speed be $u$.Given that the kinetic energy of the man is half that of the boy:$\frac{1}{2} M v^2 = \frac{1}{2} (\frac{1}{2} m u^2) = \frac{1}{2} (\frac{1}{2} (M/2) u^2) = \frac{1}{8} M u^2$.Simplifying,$v^2 = \frac{1}{4} u^2$,which implies $u = 2v$.When the man increases his speed by $1 \, m/s$,his new speed is $(v + 1)$. His new kinetic energy equals the boy's kinetic energy:$\frac{1}{2} M (v + 1)^2 = \frac{1}{2} (M/2) u^2$.Substituting $u = 2v$:$(v + 1)^2 = \frac{1}{2} (2v)^2 = \frac{1}{2} (4v^2) = 2v^2$.Taking the square root on both sides: $v + 1 = \sqrt{2} v$.Rearranging for $v$: $1 = v(\sqrt{2} - 1)$.Thus,$v = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1 \, m/s$.

$A$ man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by $1 \, m/s$ and then has the same kinetic energy as the boy. The original speed of the man was:

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