we are given
position function as
[tex]x(t)=8t-2t^2[/tex]
we know that acceleration is second derivative of position
so, firstly, we will find first derivative
[tex] x'(t)=8*1-2*2*t^1[/tex]
now, we can simplify it
[tex] x'(t)=8-4t[/tex]
now, we can find derivative again
[tex] x''(t)=0-4*1[/tex]
[tex] x''(t)=-4[/tex]
now, we can plug t=2
and we get
[tex] x''(2)=-4[/tex]
so,
the value of the person’s acceleration a at t=2s is -4..........Answer
