Integration, where am i going wrong here?

[devanlevin]

given,-

\vec{a}_{x}=6t^{2}
\vec{a}_{y}=-4

vector V(t=0)=0
Vector R(t=0)=0

1-the displacement from the 1st to the 3rd second
2-the distance traveled fron the 2st to 3rd second
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using integration,
Vx=\int(ax)dt=2t^{3}
Vy=\int(ay)dt=-4t

x=\int(Vx)dt=0.5t^{4}
y=\int(Vy)dt=-2t^{2}

then i found r(t=1)=(0.5,-4)
and r(t=3)=(40.5,-18)

delta(r)=42.38, which is the answer to question 1
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now to find the distance traveled what i would like to do is integrate the vector V from 1 to 3, but once i get an answer for x and y, how do i connect them to get the scalar D, the distance he travelled?

 

[devanlevin]

can i apply pythagarus with x and y values?

 

[physics girl phd]

That sounds right... the motion here is separable along x- and y-directions.

 

[devanlevin]

sorrry, should read

then i found r(t=1)=(0.5,-2)
and r(t=3)=(40.5,-18)

delta(r)=43.08, which is the answer to question 1,
i did this by adding the sqared values of x and y and then square root on the answer, i think this gives me the displacement

 

[devanlevin]

That sounds right... the motion here is separable along x- and y-directions.

thats fine, but what I am looking for is the total distance travelled, not the displacemetn,,, so for example, if he traveled 10m X+, 7m X-, 20mY+,10mY-, what I am looking for is 10+7+20+10, the total travelled, not sqrt(3^2+10^2),, problem here is its a complicated curve not straightforward, x then y. how do i calculate that? won't the integral just give me the displacement? also, how do i get a single scalar value, if i know that he walked 40m on X and 16 on y is the distance travelled43.08m?/ how else can i find it other thn pythagarus