Answer:
Follows are the solution to this question:
Explanation:
In point a:
Place of particles
[tex]X(t)=\int V_{x}(t)dt[/tex]
[tex]=\int 2t^{2}dt\\\\=\frac{2}{3}t^{3}+C[/tex]
[tex]\to t=0\\\\ \to X(0)=2.3 \ m[/tex]
[tex]\to X(0)=0+C\\\\ \to C=2.3\ m[/tex]
[tex]\to X(t)=( \frac{2}{3})t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( \frac{2}{3})\times 2.2^3 +2.3 \\\\[/tex]
[tex]= \frac{2}{3}\times 10.648 +2.3\\\\= \frac{21.296}{3}+2.3\\\\ = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m[/tex]
In point b:
when [tex]t=2.2 \ s[/tex]
the Particle velocity [tex](V)=2 \times 2.22 =9.68\ \frac{m}{s}[/tex]
In point c:
Calculating the Particle acceleration:
[tex]\to a=\frac{dV}{dt} =4\ t\\\\\to t=2.2 \ s\\\\\to a=4\times 2.2 =8.8 \ \frac{m}{s^2}[/tex]
