Position function x(t) question

[becksftw]

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.

 

[vela]

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

This expression for x(t) relies on the assumption that the acceleration is constant, which is not the case here.

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.

By definition, you have a(t)=v'(t) and v(t)=x'(t). If you start with a(t)=constant, you can derive the equation x(t)=x₀+v₀t+1/2at². This is probably covered in your textbook or notes.

In this problem, you want to do the same thing except start with the given expression for a(t).

 

[becksftw]

Its for a course that already expects you to remember to remember this, we're just doing the quick review chapter in the beginning. I've never taken physics though, so I hardly every deal with this stuff, so I forgot exactly how to go about the problem. So are you saying then that I want to integrate a(t)?

 

[becksftw]

*integrate a(t) twice

 

[vela]

Yes, and use the initial conditions to determine the constants of integration.