Question

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a particle moves horizontally in uniform circular motion, over a horizontal xy plane. at one instant, it moves through the point at coordinates (4.30 m, 3.70 m) with a velocity of $-4.00\hat{i}$ m/s and an acceleration of $+12.1\hat{j}$ m/s². what are the (a) x and (b) y coordinates of the center of the circular path?
(a) number unit
(b) number unit
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Explanation:

Step1: Identify velocity direction

Velocity is $\vec{v} = -4.00\hat{i}\ \text{m/s}$, so the particle moves along the negative x-axis at this instant.

Step2: Identify acceleration direction

Acceleration is $\vec{a} = +12.1\hat{j}\ \text{m/s}^2$, which points toward the center of the circle (radial acceleration for uniform circular motion). So the center lies along the positive y-direction relative to the particle's x-coordinate, and the x-coordinate of the center equals the particle's x-coordinate.
$x_{\text{center}} = x_{\text{particle}} = 4.30\ \text{m}$

Step3: Calculate radius of circular path

For uniform circular motion, $a = \frac{v^2}{r}$, solve for $r$:
$r = \frac{v^2}{a} = \frac{(-4.00)^2}{12.1} = \frac{16.0}{12.1} \approx 1.322\ \text{m}$

Step4: Find y-coordinate of center

Acceleration points in positive y-direction, so center is above the particle's y-position:
$y_{\text{center}} = y_{\text{particle}} + r = 3.70 + 1.322 \approx 5.02\ \text{m}$

Answer:

(a) $4.30\ \text{m}$
(b) $5.02\ \text{m}$