Answer:
velocity v(t) = -2cos(t)+t+2
position s(t) = [tex]-2sin(t)+\frac{t^2}{2} +2t + 3[/tex]
average of position from 2 to 5 = 17.666
Step-by-step explanation:
velocity is anti-derivative/integral of acceleration
[tex]\int{(2sin(t)+1)} \, dt[/tex] = -2cos(t)+t+C
v(0) = 0
0 = -2cos(0)+0+C = -2 +C; C = 2
v(t) = -2cos(t)+t+2
position is anti-derivative of velocity
[tex]\int{(-2cos(t)+t+2)} \, dt[/tex] = [tex]-2sin(t)+\frac{t^2}{2} +2t + C[/tex]
s(0) = 3
3 = -2sin(0) + 0 + 0+ C
3 = C
s(t) = [tex]-2sin(t)+\frac{t^2}{2} +2t + 3[/tex]
average value of a function of a domain [a, b] is given by the equation
[tex]\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx[/tex]
average of position from t = 2 to t = 5
[tex]\frac{1}{5-2} \int\limits^5_2 {(-2sin(t)+\frac{t^2}{2} +2t + 3)} \, dt[/tex]
[tex]\int\limits^5_2 {(-2sin(t)+\frac{t^2}{2} +2t + 3)} \, dt[/tex] = 2cos(5)+125/6+25+15 - 2cos(2) - 8/6 -4 + 6
=52.999 (use a calculator)
52.999/3 = 17.666
