a. In the first 4.0 seconds of its motion, the car is accelerating from rest to a certain velocity. From the velocity-versus-time graph, we can see that the velocity increases linearly with time. Therefore, we can use the equation: distance = (initial velocity + final velocity) / 2 * time to calculate the distance covered by the car in the first 4.0 seconds.
In this case, the initial velocity is 0 m/s and the final velocity is V, where V is the velocity after 4.0 s. Therefore, the distance covered by the car in the first 4.0 seconds is (0 + V) / 2 * 4 = 2V * 4.
b. In the last 2.0 seconds of its motion, the car is decelerating from a certain velocity to come to rest again. From the velocity-versus-time graph, we can see that the velocity decreases linearly with time. Therefore, we can use the equation: distance = (initial velocity + final velocity) / 2 * time to calculate the distance covered by the car in the last 2.0 seconds.
In this case, the initial velocity is V, where V is the velocity before 2.0 s and the final velocity is 0 m/s. Therefore, the distance covered by the car in the last 2.0 seconds is (V + 0) / 2 * 2 = V * 2
c. The middle portion of the car's motion corresponds to the constant velocity segment of the graph. From the graph, we can see that the velocity is a constant value for the middle portion of the motion. Therefore, we can read the value of V from the graph, it is the slope of the line from the graph, which is 22/8 = 2.75 m/s.