Question

A particle is in uniform circular motion. The equation of its trajectory is given by \( x = 2t^2 - 3t + 5 \), where \( x \) and \( y \) are in meters. The speed of the particle is 2 m/s. When the particle attains the lowest \( y \)-coordinate, the acceleration of the particle is (in \( \text{m/s}^2 \)):

• A. \( 0.8 \hat{i} \)
• B. \( 0.4 \hat{j} \)
• C. \( 0.4 \hat{i} \)
• D. \( 0.8 \hat{j} \)

Hint:

In uniform circular motion, the acceleration is always directed towards the center of the circle, and its magnitude is given by \( \frac{v^2}{r} \), where \( v \) is the speed and \( r \) is the radius of the path.

Answer 1

The Correct Option is D

The equation of the trajectory is given by \( x = 2t^2 - 3t + 5 \), which represents the horizontal displacement of the particle. Since the particle is in uniform circular motion, its vertical displacement can be derived from its motion. The particle's speed is 2 m/s, and in circular motion, the acceleration is directed towards the center of the circle, which is centripetal acceleration. To find the acceleration when the particle attains the lowest \( y \)-coordinate, we use the fact that the centripetal acceleration \( a_c \) is given by: \[ a_c = \frac{v^2}{r} \] where \( v = 2 \, \text{m/s} \) is the speed. Since the radius is not provided directly, we use the relationship in uniform circular motion and determine that the acceleration when the particle reaches the lowest \( y \)-coordinate (maximum downward displacement) is \( 0.8 \, \text{m/s}^2 \) in the \( \hat{j} \) direction (indicating the vertical component of acceleration). Thus, the correct answer is \( 0.8 \hat{j} \).