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How do you show mathematically why a small increase in your vehicle's speed results in a tremendous increase in your vehicles kinetic energy?
Wiki User
∙ 10y ago
The kinetic energy of a vehicle, of mass m kilograms and travelling at velocity v metres per second, is 1/2*m*v2.
Using calculus, d(ke)/dv = m*v.
That is, the kinetic energy (ke) increases by the product of the mass and the velocity. Whether or not that can be considered "tremendous" is another matter.
The kinetic energy of a vehicle, of mass m kilograms and travelling at velocity v metres per second, is 1/2*m*v2.
Using calculus, d(ke)/dv = m*v.
That is, the kinetic energy (ke) increases by the product of the mass and the velocity. Whether or not that can be considered "tremendous" is another matter.
The kinetic energy of a vehicle, of mass m kilograms and travelling at velocity v metres per second, is 1/2*m*v2.
Using calculus, d(ke)/dv = m*v.
That is, the kinetic energy (ke) increases by the product of the mass and the velocity. Whether or not that can be considered "tremendous" is another matter.
The kinetic energy of a vehicle, of mass m kilograms and travelling at velocity v metres per second, is 1/2*m*v2.
Using calculus, d(ke)/dv = m*v.
That is, the kinetic energy (ke) increases by the product of the mass and the velocity. Whether or not that can be considered "tremendous" is another matter.
Wiki User
∙ 10y ago
Wiki User
∙ 10y agoThe kinetic energy of a vehicle, of mass m kilograms and travelling at velocity v metres per second, is 1/2*m*v2.
Using calculus, d(ke)/dv = m*v.
That is, the kinetic energy (ke) increases by the product of the mass and the velocity. Whether or not that can be considered "tremendous" is another matter.
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the varible b can objects kinetic energy
