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?
The energy required to go from 30 mph to 60 mph is not half as much as the energy required to go from 0 to 30 mph. The relationship between kinetic energy and speed is not linear, but rather quadratic.
The kinetic energy of an object is given by the equation:
Kinetic Energy = 1/2 * mass * (velocity)^2
Since the mass of the truck remains constant, the energy required to increase its speed is directly proportional to the square of its velocity. So, if we consider the change in energy from 0 to 30 mph as E1 and the change in energy from 30 to 60 mph as E2, we can say:
E1 < E2
In other words, the energy required to go from 30 to 60 mph is greater than the energy required to go from 0 to 30 mph.
To determine the energy required to accelerate the truck from 30 mph to 60 mph compared to going from 0 mph to 30 mph, we need to understand the relationship between kinetic energy and velocity.
The kinetic energy (KE) of an object can be calculated using the formula:
KE = (1/2) * mass * velocity^2
We can assume that the mass of the truck remains constant for simplicity. So, we don't need to consider changes in mass.
Let's compare the initial and final kinetic energies:
1. From 0 to 30 mph:
Initial velocity (Vi) = 0 mph
Final velocity (Vf) = 30 mph
The change in kinetic energy (ΔKE₁) can be calculated as:
ΔKE₁ = KE at 30 mph - KE at 0 mph = (1/2) * mass * Vf^2 - (1/2) * mass * Vi^2 = (1/2) * mass * (Vf^2 - Vi^2)
2. From 30 to 60 mph:
Initial velocity (Vi) = 30 mph
Final velocity (Vf) = 60 mph
The change in kinetic energy (ΔKE₂) can be calculated as:
ΔKE₂ = KE at 60 mph - KE at 30 mph = (1/2) * mass * Vf^2 - (1/2) * mass * Vi^2 = (1/2) * mass * (Vf^2 - Vi^2)
Now, let's compare the energy required to go from 30 to 60 mph (ΔKE₂) with the energy required to go from 0 to 30 mph (ΔKE₁):
ΔKE₂ / ΔKE₁ = [(1/2) * mass * (Vf^2 - Vi^2)] / [(1/2) * mass * (Vf^2 - Vi^2)]
Notice that the mass cancels out, and we are left with:
ΔKE₂ / ΔKE₁ = (Vf^2 - Vi^2) / (Vf^2 - Vi^2)
This simplifies to:
ΔKE₂ / ΔKE₁ = 1
Therefore, the energy required to go from 30 to 60 mph (ΔKE₂) is equal to the energy required to go from 0 to 30 mph (ΔKE₁). In other words, it's not half as much; it's the same amount of energy.