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

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

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Answer 1 · 2017-03-23T06:42:29

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

$=>dot (KE)=1500*80*10$
$=>dot (KE)=1.2*10^6Js^-1$

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

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Using the equation for kinetic energy: k=1/2mv^2k=1/2mv^2 Find the instantaneous rate of change of the kinetic energy of a 1500 kg1500 kg car which has a velocity of 80 m/s80 m/s and an acceleration of 10m/s^210m/s^2?

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