Final answer:
After accelerating from rest for 8 seconds at 1.5 m/s², the car reaches a velocity of 12 m/s. When the brakes are applied for the next 5 seconds at a deceleration of -2.0 m/s², the car's final velocity becomes 2.0 m/s.
Step-by-step explanation:
The student has asked about the final velocity of a car after it has undergone two different accelerations. Initially, the car starts from rest and accelerates for 8.0 seconds at a rate of +1.5 m/s². To find the velocity at the end of this period, we use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get v = 0 + (1.5 m/s² * 8.0 s) = 12.0 m/s. This is the velocity of the car before braking.
Next, the driver applies the brakes, which causes a uniform deceleration of -2.0 m/s² for 5.0 seconds. To determine the final velocity after braking, we again use the formula v = u + at, but this time u is the velocity just before braking (12.0 m/s), a is the deceleration (-2.0 m/s²), and t is the duration of the braking (5.0 s). Therefore, the final velocity is v = 12.0 m/s + (-2.0 m/s² * 5.0 s) = 2.0 m/s. So, at the end of the braking period, the car is going at a speed of 2.0 m/s.