Final Answer:
(a) The initial acceleration of the car was 2 m/s².
(b) The velocity of the car after 7 seconds was 14 m/s.
(c) The braking acceleration of the car was -2 m/s² in the opposite direction of motion.
(d) The car braked for 5 seconds.
(e) The distance from the first traffic light to the second traffic light was 62 meters.
(f) The average velocity of the car during the entire trip was 8.8 m/s.
(g) Graph of position vs. time would show initial rest, constant acceleration, constant velocity, and then constant deceleration.
(h) Graph of velocity vs. time would show initial acceleration, constant velocity, and then constant deceleration.
(i) Graph of acceleration vs. time would show a constant positive acceleration during the initial phase and then a constant negative acceleration during braking.
Step-by-step explanation:
(a) Using the kinematic equation, we can calculate the initial acceleration: a = (2 * d) / t² = (2 * 37 m) / (7 s)² = 2 m/s².
(b) After 7 seconds, the velocity can be found using the formula: v = u + at, where u is the initial velocity (0 m/s), a is the acceleration (2 m/s²), and t is the time (7 s). So, v = 0 + (2 m/s² * 7 s) = 14 m/s.
(c) Braking acceleration is the negative of the initial acceleration, so it's -2 m/s² in the opposite direction.
(d) The car brakes until it comes to a stop, so it brakes for 5 seconds.
(e) The total distance is the sum of distances traveled during acceleration, constant speed, and deceleration: 37 m + 15 s * 14 m/s + 25 m = 62 meters.
(f) Average velocity = Total distance / Total time = 62 m / 22 s = 8.8 m/s.
(g) and (h) Graphs can be plotted based on the equations and the characteristics of motion described.
(i) The acceleration vs. time graph will show a constant positive value during acceleration and a constant negative value during deceleration, with no acceleration during the constant-speed phase.