a) On the Acme W&W company production line, parts have to be transported 100 m by conveyor belt to a machine area. The parts move at a constant speed of 2.5 m/s. How much time does it take to move a part 100 m?

b) On the same conveyor belt as described in problem a), the part must slow down to 0.2 m/s over a distance of 0.9 m once it reaches the machine area. The original speed was 2.5 m/s. What is the average acceleration while slowing down? How much time does it take to slow down?

c) When leaving the machine area described in problems a and b, the part reverses direction and starts with a velocity of 0.2 m/s. It is accelerated at 1.5 m/s^2 for its journey back a distance of 5m. How long does it take the part to travel the 5m distance? What is the final velocity of the part?

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a) To calculate the time it takes to move a part 100 m at a constant speed of 2.5 m/s, we can use the formula:

Time = Distance / Speed

Given that the distance is 100 m and the speed is 2.5 m/s, we can substitute those values into the equation:

Time = 100 m / 2.5 m/s

Simplifying:

Time = 40 seconds

Therefore, it takes 40 seconds to move a part 100 m on the conveyor belt at a constant speed of 2.5 m/s.

b) To calculate the average acceleration while slowing down and the time it takes to slow down, we can use the following equations:

Acceleration = (Final Speed - Initial Speed) / Time
Time = Distance / Average Speed

Given that the initial speed is 2.5 m/s, the final speed is 0.2 m/s, and the distance over which the part slows down is 0.9 m, we can first calculate the average speed during deceleration:

Average Speed = (Initial Speed + Final Speed) / 2
Average Speed = (2.5 m/s + 0.2 m/s) / 2
Average Speed = 1.35 m/s

Next, we calculate the time it takes to slow down:

Time = Distance / Average Speed
Time = 0.9 m / 1.35 m/s
Time = 0.67 seconds

To find the average acceleration while slowing down, we can use the first equation:

Acceleration = (Final Speed - Initial Speed) / Time
Acceleration = (0.2 m/s - 2.5 m/s) / 0.67 seconds
Acceleration = -2.3 m/s / 0.67 seconds
Acceleration ≈ -3.43 m/s^2

Therefore, the average acceleration while slowing down is approximately -3.43 m/s^2, and it takes approximately 0.67 seconds to slow down.

c) To calculate the time it takes for the part to travel the 5 m distance when leaving the machine area, we can use the equation:

Time = (Final Velocity - Initial Velocity) / Acceleration

Given that the initial velocity is 0.2 m/s and the acceleration is 1.5 m/s^2, we need to find the final velocity. We can use the equation:

Final Velocity^2 = Initial Velocity^2 + 2 * Acceleration * Distance

Final Velocity^2 = (0.2 m/s)^2 + 2 * 1.5 m/s^2 * 5 m
Final Velocity^2 = 0.04 m^2/s^2 + 15 m^2/s^2
Final Velocity^2 = 15.04 m^2/s^2

Taking the square root of both sides:

Final Velocity ≈ √15.04 m/s
Final Velocity ≈ 3.88 m/s

Now we can calculate the time:

Time = (Final Velocity - Initial Velocity) / Acceleration
Time = (3.88 m/s - 0.2 m/s) / 1.5 m/s^2
Time ≈ 2.46 seconds

Therefore, it takes approximately 2.46 seconds for the part to travel the 5 m distance when leaving the machine area, and the final velocity is approximately 3.88 m/s.