- #1
becksftw
- 8
- 0
the problem is given a(t)=4(t+3)^2, v0=-1, x0=1 find x(t), the position function
to my knowledge x(t)= 1/2at^2+v0t+x0
I tried it and got the wrong answer, the answer the book gives is: 1/3(t+3)^4 - 37t -26
I did:
x(t)= 1/2 [4(t+3)^2)]t^2 - 1t +1
eventually getting: 4t^2 + 24t + 9 for the part in parenthesis
and 2t^4 +12t^2 + 9/2t^2 after multiplying by t^2 and then 1/2
What I'm getting it shaping up to look nothing like the answer, so I know I'm missing something big here.
to my knowledge x(t)= 1/2at^2+v0t+x0
I tried it and got the wrong answer, the answer the book gives is: 1/3(t+3)^4 - 37t -26
I did:
x(t)= 1/2 [4(t+3)^2)]t^2 - 1t +1
eventually getting: 4t^2 + 24t + 9 for the part in parenthesis
and 2t^4 +12t^2 + 9/2t^2 after multiplying by t^2 and then 1/2
What I'm getting it shaping up to look nothing like the answer, so I know I'm missing something big here.