Asked by Anonymous
find the position function s(t) given acceleration a(t)=3t, if v(2)=0, and s(2)=1.
the answer is s(t)=(t^3/2)-6t +9 but I don't know how to solve this.
Can someone help me with this?
Thank you!
the answer is s(t)=(t^3/2)-6t +9 but I don't know how to solve this.
Can someone help me with this?
Thank you!
Answers
Answered by
Steve
sure you do.
a(t) = 3t
v(t) = 3/2 t^2 + c
but, v(2)=0, so
3/2(4)+c = 0
c = -6, so
v(t) = 3/2 t^2 - 6
s(t) = 1/2 t^3 - 6t + c
but, s(2)=1, so
1/2 (8) - 6(2) + c = 1
c = 9
s(t) = 1/2 t^3 - 6t + 9
a(t) = 3t
v(t) = 3/2 t^2 + c
but, v(2)=0, so
3/2(4)+c = 0
c = -6, so
v(t) = 3/2 t^2 - 6
s(t) = 1/2 t^3 - 6t + c
but, s(2)=1, so
1/2 (8) - 6(2) + c = 1
c = 9
s(t) = 1/2 t^3 - 6t + 9