To solve this problem, we'll break it down into three phases: acceleration, constant speed, and deceleration. We'll calculate the distance covered in each phase and then sum them up to find the total distance covered.
Phase 1: Acceleration The initial velocity, u = 0 (starting from rest) Final velocity, v = 70 km/h Time, t = 30 seconds
Using the formula for uniform acceleration:
v = u + at
Converting the velocities to m/s:
u = 0 m/s v = 70 km/h = 70 * (1000 m/3600 s) = 19.44 m/s
19.44 = 0 + a * 30
a = (19.44 - 0) / 30 a = 0.648 m/s² (acceleration)
To calculate the distance covered during acceleration, we can use the formula:
s = ut + (1/2) * a * t²
s = 0 * 30 + (1/2) * 0.648 * (30)² s = 0 + 9.72 s = 9.72 meters
Phase 2: Constant Speed The train maintains a constant speed of 70 km/h for 60 seconds. Since the speed is constant, the distance covered during this phase can be calculated using:
s = v * t
Converting the speed to m/s:
v = 70 km/h = 70 * (1000 m/3600 s) = 19.44 m/s t = 60 seconds
s = 19.44 * 60 s = 1166.4 meters
Phase 3: Deceleration The train decelerates uniformly from 70 km/h to rest (0 km/h) in 60 seconds. We'll use the same acceleration value calculated during the acceleration phase since the deceleration is uniform.
Using the formula:
v = u + at
Converting the velocities to m/s:
u = 19.44 m/s v = 0 m/s
0 = 19.44 + a * 60
a = (0 - 19.44) / 60 a = -0.324 m/s² (deceleration)
To calculate the distance covered during deceleration:
s = ut + (1/2) * a * t²
s = 19.44 * 60 + (1/2) * (-0.324) * (60)² s = 1166.4 - 583.2 s = 583.2 meters
Total distance covered: Adding the distances covered in each phase:
Total distance = Distance in Phase 1 + Distance in Phase 2 + Distance in Phase 3 Total distance = 9.72 + 1166.4 + 583.2 Total distance = 1759.32 meters
Therefore, the total distance covered by the diesel train is 1759.32 meters.