Let's break down the problem step by step. --- ### **Given:** - Radius of the circle, \( R = 10 \) m - Speed as a function of time: \( v(t) = 2t + 1 \) m/s - Time, \( t = 3 \) s We are to **find the particle's acceleration** at \( t = 3 \) s. --- ### **Step 1: Types of Acceleration in Circular Motion** A particle moving in a circle can have two components of acceleration: 1. **Tangential acceleration (\( a_t \))**: Due to change in speed. 2. **Centripetal (radial) acceleration (\( a_c \))**: Due to change in direction. Total acceleration, \[ a = \sqrt{a_t^2 + a_c^2} \] --- ### **Step 2: Tangential Acceleration (\( a_t \))** Tangential acceleration is the time derivative of speed: \[ a_t = \frac{dv}{dt} \] Given: \( v(t) = 2t + 1 \) \[ \frac{dv}{dt} = 2 \] So, \[ a_t = 2 \text{ m/s}^2 \] --- ### **Step 3: Centripetal Acceleration (\( a_c \))** \[ a_c = \frac{v^2}{R} \] At \( t = 3 \) s, \[ v(3) = 2(3) + 1 = 7 \text{ m/s} \] \[ a_c = \frac{(7)^2}{10} = \frac{49}{10} = 4.9 \text{ m/s}^2 \] --- ### **Step 4: Total Acceleration** \[ a = \sqrt{a_t^2 + a_c^2} \] \[ a = \sqrt{(2)^2 + (4.9)^2} \] \[ a = \sqrt{4 + 24.01} \] \[ a = \sqrt{28.01} \] \[ a \approx 5.29 \text{ m/s}^2 \] --- ### **Final Answer:** \[ \boxed{5.29\ \text{m/s}^2} \] This is the acceleration of the particle at \( t = 3 \) s.

The total acceleration of a particle moving in a circle is the vector sum of tangential acceleration, due to change in speed, and centripetal acceleration, due to change in direction.