What could be causing the discrepancy in Y0 when finding the placement vector?

[devanlevin]

given the acceleration
Ax=6t^2
Ay=-0.66PI*cos(0.66PI*t)
and knowing that the starting point was (0,0) and that the starting velocity was V0=(-1,0)
find the Velocity vector and the placement vector
si what i did was- 1st integral on Ax Ay getting
Vx= 2t^3-1
Vy=-sin(0.666PI*t)

which is fine and is true to the facts- V0=(-1,0)
but then, 2nd intergal, integral in Vx, Vy
and
Rx= 0.5t^4-t
which is fine, X0=0
BUT
Ry= 1.5PI*cos(0.66PI*t)
and then Y0 is not 0, i have checked and double checked and triple checked and cannot find my fault, any ideas?/

 

[tiny-tim]

Ay=-0.66PI*cos(0.66PI*t)
and knowing that the starting point was (0,0) and that the starting velocity was V0=(-1,0)
…
Vy=-sin(0.666PI*t)
…
Ry= 1.5PI*cos(0.66PI*t)
and then Y0 is not 0, i have checked and double checked and triple checked and cannot find my fault, any ideas?/

Hi devanlevin!

Whenever you integrate, you must include a constant of integration.

That constant just happened to be 0 in the other parts of the question …

but it isn't zero here … that's all!

 

[devanlevin]

but i did, added Vx0=-1 and Vy0=0, and isn't the constant in this case X0 and Y0, which are both 0 anyway??

 

[tiny-tim]

Hi devanlevin!

but i did, added Vx0=-1 and Vy0=0, and isn't the constant in this case X0 and Y0, which are both 0 anyway??

ah, I see now …

No, X₀ and Y₀ are the "target figures", not the constants themselves.

You have to choose a constant so that Y(0) = Y₀ …

in other words, choose C so that 1.5*π*cos(0.66πt) + C = 0.

 

[devanlevin]

thanks