a train with an initial velcity of 0.50 m/s accelerates to 2.00m/s for 2 seconds, coasts with zero acceleration for 3.0 seconds, and then accelerates at -1.5 m/s/s for 1.0 seconds
To find the distance traveled by the train, we can break down the motion into three parts: the initial acceleration, the coasting period, and the final deceleration.
1. Initial Acceleration:
The train starts from rest, so its initial velocity (u) is 0.5 m/s, and it accelerates to a final velocity (v) of 2.0 m/s over a period of 2 seconds. To find the distance covered during this time, we can use the formula:
distance = (u + v) * t / 2
distance = (0.5 + 2.0) * 2 / 2
distance = 2.5 meters
2. Coasting Period:
During the coasting period, the train maintains a constant velocity of 2.0 m/s for 3 seconds. Since the velocity remains constant, the distance covered can be calculated using the formula:
distance = velocity * time
distance = 2.0 * 3
distance = 6.0 meters
3. Final Deceleration:
In the final stage, the train decelerates at a rate of -1.5 m/s^2 (negative value indicates deceleration) for 1 second. To find the distance covered, we can use the formula:
distance = (u + v) * t / 2
distance = (2.0 + 0) * 1 / 2
distance = 1.0 meter
Total Distance Traveled = 2.5 + 6.0 + 1.0 = 9.5 meters
Therefore, the train travels a total distance of 9.5 meters.