- #1
devanlevin
given,-
[tex]\vec{a}[/tex][tex]_{x}[/tex]=6t[tex]^{2}[/tex]
[tex]\vec{a}[/tex][tex]_{y}[/tex]=-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=[tex]\int[/tex](ax)dt=2t[tex]^{3}[/tex]
Vy=[tex]\int[/tex](ay)dt=-4t
x=[tex]\int[/tex](Vx)dt=0.5t[tex]^{4}[/tex]
y=[tex]\int[/tex](Vy)dt=-2t[tex]^{2}[/tex]
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?
[tex]\vec{a}[/tex][tex]_{x}[/tex]=6t[tex]^{2}[/tex]
[tex]\vec{a}[/tex][tex]_{y}[/tex]=-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
----------------------
using integration,
Vx=[tex]\int[/tex](ax)dt=2t[tex]^{3}[/tex]
Vy=[tex]\int[/tex](ay)dt=-4t
x=[tex]\int[/tex](Vx)dt=0.5t[tex]^{4}[/tex]
y=[tex]\int[/tex](Vy)dt=-2t[tex]^{2}[/tex]
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
----------------------
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?
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