Step-by-step explanation:
1. Given: a = 4t + 4
We know that
[tex] a = \frac{dv}{dt} \: \: or \: dv = adt[/tex]
By definition, the integral of a power function x^n is
[tex] \int {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1} + k[/tex]
Integrating the acceleration a, we get
[tex]v = \int adt = \int(4t + 4)dt = 2 {t}^{2} + 4t + k[/tex]
where k = constant of integration. We know that v = 10 when t = 0 so when we do the substitution, we get k = 10 therefore, the final expression for v is
[tex]v = 2 {t}^{2} + 4t + 10[/tex]
To find s, we need to integrate v. Knowing that
[tex]v = \frac{ds}{dt} \: \: or \: s = \int v \: dt[/tex]
[tex]s = \int(2 {t}^{2} + 4t + 10)dt[/tex]
[tex] = \frac{2}{3} {t}^{3} + 2 {t}^{2} + 10t + k[/tex]
where k once again is the constant of integration. We know that s = 2 when t = 0, which gives us k = 2. Therefore, the final expression for s is
[tex]s = \frac{2}{3} {t}^{3} + 2 {t}^{2} + 10t + 2[/tex]
2. The potential difference V between two boundaries a and b is given by
[tex]V = \frac{q}{2\pi \epsilon_0 \epsilon_r} \int_{b}^{a} \frac{dr}{r} [/tex]
Note that the integral in the expression above can be rewritten and the integrated as
[tex] \int_{b}^{a} \frac{dr}{r} =\int_{b}^{a} {r}^{ - 1} dr = \ln |a| - \ln |b| [/tex]
so the potential difference V is then J
[tex]V = \frac{q}{2\pi \epsilon_0 \epsilon_r} \int_{b}^{a} \frac{dr}{r}[/tex]
[tex] = \frac{2 \times {10}^{ - 6} }{2\pi (8.85\times {10}^{ - 12}) (2.77)}( \ln |20| - \ln |10| )[/tex]
[tex] = 9000 \: V[/tex]
