Kinetic energy problem

[monkeyvokes]

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problem: The chemical potential energy in gasoline is converted to
kinetic energy in cars. If a car accelerates from zero to 60
km/h, compared to the energy necessary to increase the
velocity of the car from zero to 30 km/h, the energy
necessary to increase the velocity of the car from 30 to 60
km/h is:

the answer is 3 times as great. Why?

 

[Temperature101]

problem: The chemical potential energy in gasoline is converted to
kinetic energy in cars. If a car accelerates from zero to 60
km/h, compared to the energy necessary to increase the
velocity of the car from zero to 30 km/h, the energy
necessary to increase the velocity of the car from 30 to 60
km/h is:

the answer is 3 times as great. Why?

Dont know why my post went multiple times

 

[Temperature101]

problem: The chemical potential energy in gasoline is converted to
kinetic energy in cars. If a car accelerates from zero to 60
km/h, compared to the energy necessary to increase the
velocity of the car from zero to 30 km/h, the energy
necessary to increase the velocity of the car from 30 to 60
km/h is:
.......

 

[Temperature101]

problem: The chemical potential energy in gasoline is converted to
kinetic energy in cars. If a car accelerates from zero to 60
km/h, compared to the energy necessary to increase the
velocity of the car from zero to 30 km/h, the energy
necessary to increase the velocity of the car from 30 to 60
km/h is:

the answer is 3 times as great. Why?

Mass of the car stays the same:

First acceleration: E(total) form 0m/s to 30m/s = Ef - Ei = 0.5V^2 - 0 = 0.5*(30)^2 = 450J
2nd acceleration : E (total) from 30m/s to 60 m/s = 1850 - 450= 1350 J

 

[Vurday]

Assuming 100% efficiency:

KEf - KEi = Chemical Energy Needed
1/2m(vf2) - 1/2m(vo2) = E needed

1/2m(30^2) - 1/2m(0^2) = E 0->30 = 1/2m*900J
1/2m(60^2) - 1/2m(30^2) = E 30->60 = 1/2m*2700J

E 30->60 = 3*E 0->30

 

[Romz]

Lol I remember this from exam krackers and it still took me wayy too long to solve even now.

Tackled it the wrong way..