when a man increases his speed by `2ms^(1)`, he finds that his kinetic energy is doubled the original speed of the man is

Options

A. `2(sqrt2-1)msw^(-1)`
B. `2(sqrt2+1)ms^(-1)`
C. `4.5ms^(-1)`
D. None of these

Answer

B

Explanation

`KE = (1)/(2)mv^(2), " Given" v_(2) = (v_(1) + 2) therefore (K_(1))/(K_(2)) = ((v_(1))/(v_(2)))^(2)`
`rArr (1)/(2) = (v_(1)^(2))/((v_(1) + 2)^(2)) (therefore K_(2) = 2k_(1))`
`v_(1)^(2) + 4v_(1) + 4 = 2v_(1)^(2), " " v_(1)^(2) - 4v_(1) - 4 = 0, " " v_(1) = (4 pm sqrt(16 + 16))/(2)`
`v_(1) = (4 + sqrt(32))/(2) = 2 (sqrt(2 ) + 1)ms^(-1)`