Answer:
The position of the particle as a function of times is given by
[tex]x(t)=t^{2}+2t+x(o)[/tex] where [tex]x(0)[/tex] is the position of the particle at t = 0 secs.
Step-by-step explanation:
Give that
[tex]v(t)=2t+4[/tex]
we know that
[tex]v(t)=\frac{dx(t)}{dt}\\\\\therefore dx(t)=v(t)dt\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt[/tex]
Applying values we get
[tex]x(t)=\int (2t+4)dt\\\\x(t)=\int 2tdt+\int 4dt\\\\x(t)=\frac{2t^{2}}{2}+4t+c[/tex]
here 'c' is the constant of integration which represents the position of the particle at time t = 0 secs.
Thus we have
[tex]x(t)=t^{2}+2t+x(o)[/tex]
