Answer:
Position at t= 4 seconds is 144 m
Explanation:
It is given that acceleration, a = 18 t, where t is the time.
We know that Velocity, [tex]v = \int { a} \, dt[/tex]
Substituting value of a,
Velocity, [tex]v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c[/tex]
We know that at t = 0, v = -12 m/s
So, [tex]9*0^2+c=-12\\ \\ c=-12m/s[/tex]
So velocity, [tex]v = (9t^2-12)m/s[/tex]
We also know that displacement, [tex]x = \int { v} \, dt[/tex]
Substituting value of v,
Displacement, [tex]x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c[/tex]
We know that at t = 0, particle is at origin, x =0.
So, [tex]0=3*0^3-12*0+c\\ \\ c=0[/tex]
Displacement, [tex]x = 3t^3-12t[/tex]
At t = 4 seconds
[tex]x = 3*4^3-12*4=192--48=144m[/tex]
Position at t= 4 seconds is 144 m
