a train starts from rest and accelerates uniformly to achieve a velocity of 20m/sec in 10 sec.then train maintains this speed for next 200 sec.the breaks are then applied and train comes to rest in next 50 sec.calculate
a)acceleration in first 10 sec.
b)acceleration in last 50 sec.
c)the total distance traveled by the train in whole journey
d)average velocity of the trip.
See 5-28-12, 3:02 am post.
To calculate the answers to these questions, we will use the equations of motion.
a) Acceleration in the first 10 seconds:
We can use the equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given that the initial velocity, u, is 0 m/s (since the train starts from rest) and the final velocity, v, is 20 m/s, and the time, t, is 10 seconds, we can rearrange the equation to solve for acceleration, a.
a = (v - u) / t
a = (20 m/s - 0 m/s) / 10 s
a = 2 m/s²
So, the acceleration in the first 10 seconds is 2 m/s².
b) Acceleration in the last 50 seconds:
Since the train maintains a constant speed of 20 m/s for the next 200 seconds, the acceleration during this period is 0 m/s². Therefore, the acceleration in the last 50 seconds, when the brakes are applied, is also 0 m/s².
c) Total distance traveled by the train in the whole journey:
To calculate the total distance traveled, we need to consider the distance covered during the acceleration phase, the distance covered during the constant speed phase, and the distance covered during the deceleration phase.
During the acceleration phase, we can use the equation of motion, s = ut + (1/2)at², where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the initial velocity, u, is 0 m/s, the acceleration, a, is 2 m/s², and the time, t, is 10 seconds, we can calculate the distance traveled during the acceleration phase:
s = (1/2)at²
s = (1/2) * 2 m/s² * (10 s)²
s = 2 m/s² * 100 s²
s = 200 meters
During the constant speed phase, the time is 200 seconds, and the speed is 20 m/s. So, the distance traveled during this phase is given by:
s = vt
s = 20 m/s * 200 s
s = 4000 meters
During the deceleration phase, the final velocity is 0 m/s, the acceleration is 0 m/s², and the time is 50 seconds. Therefore, the distance traveled during this phase is 0 meters.
The total distance traveled is the sum of the distances covered during each phase:
Total distance = distance during acceleration + distance during constant speed + distance during deceleration
Total distance = 200 meters + 4000 meters + 0 meters
Total distance = 4200 meters
So, the total distance traveled by the train in the whole journey is 4200 meters.
d) Average velocity of the trip:
Average velocity is defined as the total displacement divided by the total time taken.
The total displacement is the change in position from the initial position to the final position, which is 4200 meters.
The total time taken can be calculated by adding the times taken during each phase:
Total time = time taken during acceleration + time taken during constant speed + time taken during deceleration
Total time = 10 seconds + 200 seconds + 50 seconds
Total time = 260 seconds
Average velocity = total displacement / total time
Average velocity = 4200 meters / 260 seconds
Average velocity ≈ 16.15 m/s
So, the average velocity of the trip is approximately 16.15 m/s.