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A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.7 seconds, coasts for 2.1 s , and then slows down at a rate of 1.5m/s2 for the next stop sign.

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Answer:

Step-by-step explanation:

To analyze the motion of the car, we can break down the given information into different phases: acceleration, coasting, and deceleration.

1. Acceleration phase:

The car accelerates at a rate of 2.0 m/s^2 for a duration of 6.7 seconds. This means that every second, the car's velocity increases by 2.0 m/s. To find the final velocity at the end of this phase, we can use the equation:

final velocity = initial velocity + (acceleration * time)

Since the car starts from rest (initial velocity = 0), the final velocity after 6.7 seconds of acceleration would be:

final velocity = 0 + (2.0 * 6.7) = 13.4 m/s.

2. Coasting phase:

After the acceleration phase, the car coasts for 2.1 seconds. During this phase, the car maintains a constant velocity. Therefore, the velocity remains at 13.4 m/s throughout the coasting phase.

3. Deceleration phase:

The car then slows down at a rate of 1.5 m/s^2. To find the time it takes for the car to come to a stop, we can use the equation:

final velocity = initial velocity + (acceleration * time)

Since the final velocity at the end of this phase is 0, and the initial velocity is 13.4 m/s, we can solve for the time it takes to decelerate to a stop:

0 = 13.4 + (-1.5 * time)

-13.4 = -1.5 * time

time = 8.93 seconds.

Therefore, the car takes a total of 6.7 seconds for acceleration, 2.1 seconds for coasting, and 8.93 seconds for deceleration to come to a stop at the next stop sign.

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