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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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