If a man increase his speed by $2 \,m/s$ , his K.E. is doubled, the original speed of the man is
If a man increase his speed by $2 \,m/s$ , his K.E. is doubled, the original speed of the man is
$(1 + 2\sqrt 2 )\;m/s$
$4\, m/s$
$(2 + 2\sqrt 2 )\,m/s$
$(2 + \sqrt 2 )\;m/s$
If a man increase his speed by $2 \,m/s$ , his K.E. is doubled, the original speed of the man is
Initial kinetic energy $E = \frac{1}{2}m{v^2}$ …(i)
Final kinetic energy $2E = \frac{1}{2}m{(v + 2)^2}$…(ii)
by solving equation (i) and (ii) we get
$v = (2 + 2\sqrt 2 )\;m/s$