- #36
Hootenanny
Staff Emeritus
Science Advisor
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Saladsamurai said:Oh man...
Is it [tex]v(t)=\frac{d\vec{r}}{dt}=[\frac{dx}{dt}\hat i+2x\frac{dx}{dt}\hat j]\frac{m}{s}[/tex]?
Correct! And can you now find the acceleration...?
Aside:
It is common practice in physics and applied mathematics to use Newton's notation to make things a little easier to read;
[tex]\frac{dx}{dt} = \dot{x}\hspace{1cm};\hspace{1cm}\frac{d^2x}{dt^2} = \ddot{x}(t)[/tex]
So your velocity would become;
[tex]\underline{v}(t) = \underline{\dot{r}}(t)= \dot{x}(t)\hat{i}+ 2x\cdot\dot{x}(t)\hat{j}[/tex]
But either notation is equally correct, so just use whichever you feel most comfortable with.
[tex]\frac{dx}{dt} = \dot{x}\hspace{1cm};\hspace{1cm}\frac{d^2x}{dt^2} = \ddot{x}(t)[/tex]
So your velocity would become;
[tex]\underline{v}(t) = \underline{\dot{r}}(t)= \dot{x}(t)\hat{i}+ 2x\cdot\dot{x}(t)\hat{j}[/tex]
But either notation is equally correct, so just use whichever you feel most comfortable with.
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