What force is required to bring a 1200 kg car to rest from \( 22 \mathrm{~m} / \mathrm{s} \) in 3 seconds?

To determine the force required to bring a 1200 kg car to rest from a speed of \( 22 \, \text{m/s} \) in 3 seconds, we can use the following steps: 1. **Calculate the acceleration (deceleration):** The formula for acceleration \( a \) is: \[ a = \frac{\Delta v}{\Delta t} \] where: - \( \Delta v \) is the change in velocity. - \( \Delta t \) is the change in time. Here, the car is coming to rest, so the final velocity \( v_f = 0 \, \text{m/s} \) and the initial velocity \( v_i = 22 \, \text{m/s} \). \[ a = \frac{0 - 22 \, \text{m/s}}{3 \, \text{s}} = \frac{-22}{3} \, \text{m/s}^2 \approx -7.33 \, \text{m/s}^2 \] The negative sign indicates that the acceleration is in the opposite direction of the motion (deceleration). 2. **Calculate the force:** Using Newton's second law: \[ F = m \times a \] where: - \( m = 1200 \, \text{kg} \) - \( a = -7.33 \, \text{m/s}^2 \) \[ F = 1200 \, \text{kg} \times (-7.33 \, \text{m/s}^2) = -8800 \, \text{N} \] The negative sign indicates that the force is applied in the direction opposite to the car's motion. 3. **Magnitude of the Force:** Since force is a vector, if you're interested in the magnitude: \[ |F| = 8800 \, \text{N} \] **Final Answer:** A force of approximately 8.8 × 10³ newtons is required to bring the car to rest.