The chemical potential energy of a certain amount of gasoline is converted into kinetic energy in a truck that increases its speed from 0 mph to 30 mph. To pass another truck the driver accelerates from 30 mph to 60 mph. Compared to the energy required to go from 0 to 30 mph, the energy required to go from 30 to 60 mph is?

Question

The chemical potential energy of a certain amount of gasoline is converted into kinetic energy in a truck that increases its speed from 0 mph to 30 mph. To pass another truck the driver accelerates from 30 mph to 60 mph. Compared to the energy required to go from 0 to 30 mph, the energy required to go from 30 to 60 mph is?

I say half as much?

Answer 1

use Energy = 0.5 mv^2

(this should use SI units but the conversion factors will disappear in the ratio)

at 0 the energy is 0

at 30 the energy is =0.5xmx30^2

the enrgy required to go from 0 to 30 is 0.5xmx30^2 - 0 = 0.5xmx30^2

at 60 the energy is = 0.5xmx60^2

the enrgy required to go from 30 to 60
is 0.5xmx60^2 - 0.5xmx30^2

which is 0.5xm(60^2 - 30^2)

so the ratio is

0.5xm(60^2 - 30^2)/0.5xmx30^2

= (60^2 - 30^2)/30^2

times as much

Answer 2

To determine the energy required to accelerate from 30 mph to 60 mph compared to the energy required to accelerate from 0 to 30 mph, we need to understand the relationship between kinetic energy and velocity.

The kinetic energy of an object is given by the equation KE = 0.5 * m * v^2, where KE is the kinetic energy, m is the mass of the object, and v is the velocity.

Since we are considering only the energy required for acceleration, we can assume that the mass of the truck remains constant.

Let's break down the problem into two parts:

1. Energy required to go from 0 to 30 mph:

Let's assume the initial kinetic energy of the truck is KE1, and the initial velocity is 0 mph. The final velocity is 30 mph.

Therefore, the numerical value of the initial kinetic energy, KE1, is 0.

The kinetic energy equation becomes KE1 = 0.5 * m * (30^2) = 0.5 * m * 900 = 450 * m.

2. Energy required to go from 30 to 60 mph:

Let's assume the kinetic energy required for this part is KE2. In this case, the initial velocity is 30 mph, and the final velocity is 60 mph.

Now, the kinetic energy equation becomes KE2 = 0.5 * m * (60^2) = 0.5 * m * 3600 = 1800 * m.

Comparing the two energies, we have KE2 = 1800 * m and KE1 = 450 * m.

Therefore, the energy required to go from 30 to 60 mph (KE2) is four times greater than the energy required to go from 0 to 30 mph (KE1), not half as much.

Hence, you could conclude that the energy required to go from 30 to 60 mph is four times the energy required to go from 0 to 30 mph.