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# Using the equation for kinetic energy: $$k = \dfrac{1}{2}mv^{2}$$ Find the instantaneous rate of change of the kinetic energy of a 1500kg car which has a velocity of $$80 \dfrac{m}{s}$$ and an acceleration of $$10\dfrac{m}{s^{2}}$$ ?

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#### Answer:Kinetic energy $$KE = \frac{1}{2}mv^{2}$$Instantaneous rate of change of KE is its differential with time$$KE = \dfrac{d}{dt} \left ( \dfrac{1}{2} mv^{2} \right )$$$$\Rightarrow KE = \dfrac{1}{2}m . 2v . \dfrac{dv}{dt}$$$$\Rightarrow (KE) = m . v . a$$Where .v=a, acceleration.Inserting given instantaneous values we get$$\Rightarrow KE = 1500\times 80\times 10$$$$\Rightarrow KE = 1.2\times10^{6} Js^{-1}$$

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