A train upon passing point A at a speed of 72 kph accelerates at 0.75 m/s^2 for one minute along a straight path then decelerates at 1.0 m/s^2. How far in kilometers from point A will it be in 2 minutes after passing point A.
To solve this problem, we need to break it down into three parts based on the acceleration and deceleration phases:
1. First Phase: Acceleration at 0.75 m/s^2 for one minute
Distance covered in the first minute:
v1 = 72 kph = 20 m/s (Since 1 kph = 1/3.6 m/s)
a = 0.75 m/s^2
t = 1 minute = 60 seconds
Using the formula:
d1 = v1t + 0.5at^2
d1 = 20 * 60 + 0.5 * 0.75 * 60^2
d1 = 1200 + 1350
d1 = 2550 meters
2. Second Phase: Constant speed for one minute
During the second minute, the train is moving at a constant speed of 72 kph for 1 minute, covering a distance of:
d2 = v1 * t
d2 = 20 * 60
d2 = 1200 meters
3. Third Phase: Deceleration at 1.0 m/s^2 for one minute
During the third minute, the train starts decelerating at 1.0 m/s^2 for 1 minute:
a = -1.0 m/s^2 (deceleration)
t = 1 minute = 60 seconds
Using the formula:
d3 = v2t + 0.5at^2
d3 = 20 * 60 + 0.5 * (-1.0) * 60^2
d3 = 1200 - 1800
d3 = -600 meters
Total distance covered in 2 minutes after passing point A:
Total distance = d1 + d2 + d3
Total distance = 2550 + 1200 + (-600)
Total distance = 3150 meters
Converting the total distance to kilometers:
Total distance in kilometers = 3150 meters / 1000 = 3.15 km
Therefore, the train will be 3.15 kilometers from point A after 2 minutes of passing it.