An object moves along the \( x \) axis according to the equa- tion \( x=\left(3.00 t^{2}-2.00 t+3.00\right) \mathrm{m} \). Determine (a) the average speed between \( t=2.00 \mathrm{~s} \) and \( t=3.00 \mathrm{~s} \), (b) the instantaneous speed at \( t=2.00 \mathrm{~s} \) and at \( t= \) 3.00 s, (c) the average acceleration between \( t=2.00 \mathrm{~s} \) and \( t=3.00 \mathrm{~s} \), and (d) the instantaneous acceleration at \( t=2.00 \mathrm{~s} \) and \( t=3.00 \mathrm{~s} \).

To solve the problem, we will use the given position function \( x(t) = 3.00 t^2 - 2.00 t + 3.00 \) m. ### (a) Average Speed between \( t=2.00 \, \text{s} \) and \( t=3.00 \, \text{s} \) The average speed is defined as the total distance traveled divided by the total time taken. We first need to find the positions at \( t=2.00 \, \text{s} \) and \( t=3.00 \, \text{s} \). 1. Calculate \( x(2.00) \): \[ x(2.00) = 3.00(2.00)^2 - 2.00(2.00) + 3.00 = 3.00(4) - 4 + 3.00 = 12.00 - 4 + 3.00 = 11.00 \, \text{m} \] 2. Calculate \( x(3.00) \): \[ x(3.00) = 3.00(3.00)^2 - 2.00(3.00) + 3.00 = 3.00(9) - 6 + 3.00 = 27.00 - 6 + 3.00 = 24.00 \, \text{m} \] 3. Calculate the average speed: \[ \text{Average Speed} = \frac{x(3.00) - x(2.00)}{3.00 - 2.00} = \frac{24.00 \, \text{m} - 11.00 \, \text{m}}{1.00 \, \text{s}} = \frac{13.00 \, \text{m}}{1.00 \, \text{s}} = 13.00 \, \text{m/s} \] ### (b) Instantaneous Speed at \( t=2.00 \, \text{s} \) and \( t=3.00 \, \text{s} \) The instantaneous speed is given by the magnitude of the velocity, which is the derivative of the position function with respect to time. 1. Find the velocity function \( v(t) \): \[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(3.00 t^2 - 2.00 t + 3.00) = 6.00 t - 2.00 \] 2. Calculate \( v(2.00) \): \[ v(2.00) = 6.00(2.00) - 2.00 = 12.00 - 2.00 = 10.00 \, \text{m/s} \] 3. Calculate \( v(3.00) \): \[ v(3.00) = 6.00(3.00) - 2.00 = 18.00 - 2.00 = 16.00 \, \text{m/s} \] ### (c) Average Acceleration between \( t=2.00 \, \text{s} \) and \( t=3.00 \, \text{s} \) The average acceleration is defined as the change in velocity divided by the change in time. 1. Calculate the average acceleration: \[ \text{Average Acceleration} = \frac{v(3.00) - v(2.00)}{3.00 - 2.00} = \frac{16.00 \, \text{m/s} - 10.00 \, \text{m/s}}{1.00 \, \text{s}} = \frac{6.00 \, \text{m/s}}{1.00 \, \text{s}} = 6.00 \, \text{m/s}^2 \] ### (d) Instantaneous Acceleration at \( t=2.00 \, \text{s} \) and \( t=3.00 \, \text{s} \) The instantaneous acceleration is the derivative of the velocity function. 1. Find the acceleration function \( a(t) \): \[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6.00 t - 2.00) = 6.00 \, \text{m/s}^2 \] 2. Since the acceleration is constant, we have: \[ a(2