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Graph is position x position. There are 3 points, $A$, $B$ and $C$.

  • $A(0,2)$
  • $B(4,2)$
  • $C(6,0)$

Particle travels from $A$ to $B$ and from $B$ to $C$ at constant $v = 2 ~m/s$.

Find vector position in time for the $BC$ part of the motion?

For the AB part I did this: $v$ is constant, so position in time is given by $2\hat{i}\cdot \mathrm{t} + 2\hat{j}$. Velocity in $i$ direction x time, plus the constant $2\hat{j}$ vector component.

For BC the "angled" displacement vector is making things a bit harder. Displacement vector has two components, $2\hat{i} + (-2\hat{j})$. Absolute value of it is $2\sqrt{2}$.

Tried to decompose the velocity vector in two: $\hat{i}\sqrt{2}$ $-\hat{j}\sqrt{2}$. Then integrate each component and add up the two. But the result is not quite right...

Integrated each component independently: $\hat{i}\mathrm{t}\cdot\sqrt{2} + c$ and $-\hat{j}\mathrm{t}\cdot\sqrt{2} + c$

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2 Answers 2

I suggest that first you find the time particle needs for AB part ($t_1$) and BC part ($t_2$). $t_1$ is starting time for the second part.

For the second part, you must divide displacement with the $t_2$ to obtain speed. Only then you can use $\vec{r} = \vec{v} (t-t_1) + \vec{r}_B$.

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How about fitting a circle though the three points $A$, $B$ and $C$ and having the particle move about the circle with constant speed.

Circle equation is $(x-2)^2+(y+2)^2=20$ or center at $(2,-2)$ and radius $r=2\sqrt{5}$.

This makes the angular velocity equal to $\omega = \frac{v}{r} = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}}$

The path is then $$x(t) = \left(-2\cos(\omega t)+4\sin(\omega t)+2 \right)$$ $$y(t) = \left(4\cos(\omega t)+2 \sin(\omega t)-2 \right)$$

with $t=0\ldots\frac{5\pi}{2}$


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