0
$\begingroup$

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 $-\hat{j}\mathrm{t}\cdot\sqrt{2} + c$

$\endgroup$
1
$\begingroup$

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 $\vec{r} = \vec{v} (t-t_1) + \vec{r}_B$.

$\endgroup$
0
$\begingroup$

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

Geogebra

$\endgroup$
  • $\begingroup$ This is cool, but its unnecessary $\endgroup$ – Bryson S. Sep 14 '14 at 0:17
  • $\begingroup$ This is to illustrate that curve fitting without an underlying physical model is useless. The OP needs to establish a physical model first before trying to interpolate between points A B and C $\endgroup$ – ja72 Sep 14 '14 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.