When the speed of a vehicle increases by 2 m | vedclass

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Question

(C) Let the original speed of the vehicle be $v$.The kinetic energy $K_1 = \frac{1}{2}mv^2$.When the speed increases by $2 \ m/s$,the new speed is $(v

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the original speed of the vehicle be $v$.The kinetic energy $K_1 = \frac{1}{2}mv^2$.When the speed increases by $2 \ m/s$,the new speed is $(v + 2) \ m/s$.The new kinetic energy $K_2 = \frac{1}{2}m(v + 2)^2$.According to the problem,$K_2 = 2K_1$.Therefore,$\frac{1}{2}m(v + 2)^2 = 2 	imes (\frac{1}{2}mv^2)$.$(v + 2)^2 = 2v^2$.Taking the square root on both sides: $v + 2 = \sqrt{2}v$.Rearranging the terms: $2 = \sqrt{2}v - v = v(\sqrt{2} - 1)$.$v = \frac{2}{\sqrt{2} - 1}$.Rationalizing the denominator: $v = \frac{2(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2(\sqrt{2} + 1)}{2 - 1} = 2(\sqrt{2} + 1) \ m/s$.

When the speed of a vehicle increases by $2 \ m/s$,its kinetic energy doubles. What is the original speed of the vehicle?

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