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

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

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Answer 1 · 2017-03-23T06:42:29

Kinetic energy $K E = \frac{1}{2} m v^{2}$ $KE =1/2mv^2$
Instantaneous rate of change of KE is its differential with time
$. K E = \frac{d}{d t} (\frac{1}{2} m v^{2})$ $dot (KE)=d/dt (1/2mv^2)$
$\Rightarrow . K E = \frac{1}{2} m \cdot 2 v \cdot \frac{d v}{d t}$ $=>dot (KE)=1/2m*2v* (dv)/dt$
$\Rightarrow (. K E) = m \cdot v \cdot a$ $=>(dot KE)=m*v*a$
Where $. v = a$ $dot v=a$ , acceleration.
Inserting given instantaneous values we get

$\Rightarrow . K E = 1500 \cdot 80 \cdot 10$ $=>dot (KE)=1500*80*10$
$\Rightarrow . K E = 1.2 \cdot 10^{6} J s^{- 1}$ $=>dot (KE)=1.2*10^6Js^-1$

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

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