question_answer

If the speed of a vehicle increases by 2 m/s, its kinetic energy is doubled. What was the original speed of the vehicle?

Answer:

               

Case-1 (before)	Case-2 (after)
\[{{v}_{1}}=x\,\,m/s\]	\[{{v}_{2}}=(x+2)m/s\]
\[K.E{{.}_{1}}=E\]	\[K.E{{.}_{2}}=2E\]

We know, \[K.E.=\frac{1}{2}m{{v}^{2}}\] \[\therefore E=\frac{1}{2}m{{x}^{2}}\]                                 ????.(1) \[2E=\frac{1}{2}m{{(x+2)}^{2}}\]                               ????.(2) Dividing (1) and (2), we get \[\frac{2E}{E}=\frac{\frac{1}{2}m{{(x+2)}^{2}}}{\frac{1}{2}m{{x}^{2}}}\] \[\Rightarrow 2=\frac{{{(x+2)}^{2}}}{{{x}^{2}}}\Rightarrow 2{{x}^{2}}={{x}^{2}}+4x+4\] \[\Rightarrow {{x}^{2}}-4x-4=0\Rightarrow {{(x-2)}^{2}}-8=0\] \[\Rightarrow {{(x-2)}^{2}}=8\Rightarrow (x-2)=\sqrt{8}\] \[\Rightarrow x=\sqrt{8}+2=2.828+2=4.828\,m/s\]