### Motion and Deceleration in VehiclesA motorist traveling at 45 km/h applies the brake so that the car decelerates uniformly at 4.0 m/s². Below are several questions concerning this scenario:#### a. How far does the car travel during the 2.0 sec immediately after the brakes are applied?To determine the distance traveled by the car during the first 2.0 seconds after brakes are applied, we can use the formula:\[ \text{Distance} = v_i t + \frac{1}{2} a t^2 \]where:- \( v_i \) is the initial velocity,- \( t \) is the time,- \( a \) is the acceleration.First, convert the initial velocity from km/h to m/s:\[ v_i = 45 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 12.5 \, \text{m/s} \]Now, apply the values to the formula:\[ d = (12.5 \, \text{m/s} \times 2.0 \, \text{s}) + \frac{1}{2} (-4.0 \, \text{m/s}^2 \times (2.0 \, \text{s})^2 ) \]\[ d = 25.0 \, \text{m} + \frac{1}{2} (-16.0 \, \text{m}) \]\[ d = 25.0 \, \text{m} - 8.0 \, \text{m} \]\[ d = 17.0 \, \text{m} \]So, the car travels **17.0 meters** in 2.0 seconds.#### b. Find the car’s velocity at the end of this 2.0-sec time interval?To find the velocity at the end of the 2.0-second interval, use the equation:\[ v_f = v_i + a t \]where:- \( v_f \) is the final velocity,- \( v_i \) is the initial velocity (12.5 m/s),- \( a \) is the acceleration (-

Name: ### Motion and Deceleration in VehiclesA motorist traveling at 45 km/
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: ### Motion and Deceleration in VehiclesA motorist traveling at 45 km/h applies the brake so that the car decelerates uniformly at 4.0 m/s². Below are several questions concerning this scenario:#### a. How far does the car travel during the 2.0 sec i