91Ó°ÊÓ

1. The coordinates of point P( 4.00 m, 4.00 m)
2. The velocity is, $\stackrel{→}{v}=-5.00\hat{i}m/s$
3. The acceleration is $\stackrel{→}{a}=+12.5\hat{j}m/s^2$

Using the equation for acceleration, we can find the radius of the circular path. The particle is moving with uniform circular motion. Hence, the magnitude of velocity and acceleration is constant. The acceleration of the particle is positive and going towards the center. When we draw the figure, the coordinates of the center of the circular path will be (x,y+r).

Formula:

$a=\frac{v^2}{r}$

The particle is moving in a circular path with uniform circular motion, hence the centripetal acceleration is

$$\begin{gathered} a=\frac{v^2}{r} \\ r=\frac{v^2}{a} \\ =\frac{{(-5.00m/s)}^2}{12.5m/s^2} \\ =2.00m \end{gathered}$$

From the figure, let P( 4.00 m, 4.00 m ) be any point on the circular path, C be the center of the circle, and its coordinates will be C ( x, y + r ). From the figure, x coordinate is the same as P point.

Therefore, the x coordinate of the center of the circular path is 4.00 m .

The acceleration is positive, so it is intheupward direction at that point. Then the y coordinate of the center will be the y coordinate of P plus the circle’s radius.

$$\begin{gathered} \mathrm{C}(\mathrm{x},\mathrm{y}+\mathrm{r})=\mathrm{C}(4.00\mathrm{m},6.00\mathrm{m}) \\ \mathrm{y}=6.00\mathrm{m} \end{gathered}$$

Therefore, the y coordinate of the center of the circular path is 6.00 m .