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Let 
be the average acceleration over the first 2.46 seconds, and 
the average acceleration over the next 6.79 seconds.
At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let 
be the velocity of the car after the first 2.46 seconds.
By definition of average acceleration, we have
and we're also told that
(or possibly the other way around; I'll consider that case later). We can solve for in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:
Now we can solve for . We find that
In the case that the ratio of accelerations is actually
we would instead have
in which case we would get a velocity of
Answer:
U₁ = (ϵAV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
Explanation:
The energy stored in a capacitor is given by (1/2) (CV²)
Energy in the capacitor initially
U = CV²/2
V = voltage across the plates of the capacitor
C = capacitance of the capacitor
But the capacitance of a capacitor depends on the geometry of the capacitor is given by
C = ϵA/d
ϵ = Absolute permissivity of the dielectric material
A = Cross sectional Area of the capacitor
d = separation between the capacitor
So,
U = CV²/2
Substituting for C
U = ϵAV²/2d
Now, for U₁, the new distance between plates, d₁ = 3d
U₁ = ϵAV²/2d₁
U₁ = ϵAV²/(2(3d))
U₁ = (ϵAV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
Ohm's law states that V = IR
87 = 2 x R
R = 87/2 ohms
Hope this helps :)
