Q. If a man increase his speed by $2\ m/s$, his K.E. is doubled, the original speed of the man is
A) $\left( 1 + 2\sqrt{2} \right)\ m/s$
B) $4\ m/s$
C) $\left( 2 + 2\sqrt{2} \right)\ m/s$
D) $\left( 2 + \sqrt{2} \right)\ m/s$
Explanation
(c) Initial kinetic energy $E = \frac{1}{2}mv^{2}$ …(i)
Final kinetic energy $2E = \frac{1}{2}m(v + 2)^{2}$ …(ii)
By solving equation (i) and (ii) we get
$v = \left( 2 + 2\sqrt{2} \right)\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.