nyy Barbarino is driving his classic 1965 corvette stingray southward at \( 15.75 \mathrm{~m} / \mathrm{s} \), when he eeds up to \( 22.9 \mathrm{~m} / \mathrm{s} \). If the car's mass is \( 1,435 \mathrm{Kg} \), what is the change in momentum for s action?

To find the change in momentum, we first need to calculate the initial and final momentum of the car. The formula for momentum (p) is given by: \[ p = m \cdot v \] where \( m \) is the mass and \( v \) is the velocity. 1. **Initial momentum (p_initial)**: \[ p_{\text{initial}} = m \cdot v_{\text{initial}} = 1,435 \, \text{kg} \cdot 15.75 \, \text{m/s} = 22,593.75 \, \text{kg m/s} \] 2. **Final momentum (p_final)**: \[ p_{\text{final}} = m \cdot v_{\text{final}} = 1,435 \, \text{kg} \cdot 22.9 \, \text{m/s} = 32,825.5 \, \text{kg m/s} \] 3. **Change in momentum (\( \Delta p \))**: \[ \Delta p = p_{\text{final}} - p_{\text{initial}} = 32,825.5 \, \text{kg m/s} - 22,593.75 \, \text{kg m/s} = 10,231.75 \, \text{kg m/s} \] So, the change in momentum for Barbarino's acceleration is approximately \( 10,231.75 \, \text{kg m/s} \).