a train starts from rest and accelerates uniformly at 100m/min^2 for 10 min.then it maintains a constant velocity for remaining 20 min.brakes are applied and train uniformly retards and comes to rest in next 5 min.
a)the maximum velocity reached
b)the retardation in last 5 min
c)total distance traveled
d)the average velocity of the train.
v=a•t1 =100•10 = 1000 m/min.
s1 = a1•t1²/2 =100•100/2 =5000 m.
s2=v•t2 =1000•20 =20000 m.
0 =v +a2•t3,
a2 = -v/t3 = - 1000/5 = - 200 m/min,
s3 = v •t3 – a2•t3²/2 =1000•5 - 200•25/2 =2500 m
s = s1 + s2 +s3 = 5000 + 20000+ 2500 = 27500 m.
t =t1+t2+t3 = 10 +20 =5 = 35 min.
v(ave) =s/t = 27500/35 =785 m/min.
a) 1000 m/min,
b) - 200 m/min,
c) s = 27500 m.
d) v(ave) =785 m/min.
To solve this problem, we'll use a few equations of motion.
a) The maximum velocity reached:
We know that the train starts from rest, accelerates uniformly, and maintains a constant velocity. The formula to calculate the final velocity (v) is:
v = u + at
Where:
- v is the final velocity
- u is the initial velocity (assumed to be zero because the train starts from rest)
- a is the acceleration
- t is the time
Given that the acceleration, a, is 100m/min^2 and the time, t, is 10 minutes, we can substitute these values into the equation:
v = 0 + 100 * 10
v = 1000 m/min
Therefore, the maximum velocity reached by the train is 1000 m/min.
b) The retardation in the last 5 minutes:
In the last 5 minutes, the train applies the brakes and experiences uniform retardation. The formula to calculate the retardation (a) is:
a = (v - u) / t
Where:
- a is the retardation
- v is the final velocity (assumed to be zero because the train comes to rest)
- u is the initial velocity
- t is the time
Given that the final velocity, v, is zero and the time, t, is 5 minutes, we can substitute these values into the equation:
0 = (u - 0) / 5
0 = u / 5
u = 0
Therefore, the retardation in the last 5 minutes is zero. This means the train comes to rest gradually without any retardation.
c) The total distance traveled:
To calculate the total distance traveled, we'll use the equation for distance (s) given uniform acceleration:
s = ut + (1/2)at^2
For the first phase, where the train accelerates uniformly, the time, t, is 10 minutes. The initial velocity, u, is zero. So we have:
s1 = 0 * 10 + (1/2) * 100 * (10)^2
s1 = 0 + 5000
s1 = 5000 m
For the second phase, where the train maintains a constant velocity, the time, t, is 20 minutes. The velocity, v, is constant at 1000 m/min. So we have:
s2 = 1000 * 20
s2 = 20,000 m
For the third phase, where the train s and comes to rest, the time, t, is 5 minutes. The initial velocity, u, is 1000 m/min.
s3 = 1000 * 5 - (1/2) * a * (5)^2
s3 = 5000 - (1/2) * a * 25
Since the train comes to rest, the final distance, s, should be equal to the sum of s1, s2, and s3. Hence:
s3 = (5000 + 20,000) + s3
s3 = 25,000 m + s3
Now, using the equation for the retardation in the last 5 minutes:
0 = u - a * t
0 = 1000 - a * 5
a * 5 = 1000
a = 200 m/min^2
Substituting this value back into s3:
s3 = 25,000 m + (5000 - (1/2) * 200 * 25)
s3 = 25,000 m + (5000 - 2500)
s3 = 25,000 m + 2500
s3 = 27,500 m
Therefore, the total distance traveled by the train is 27,500 meters.
d) The average velocity of the train:
To calculate the average velocity, we'll use the formula:
Average velocity = Total distance traveled / Total time taken
We've already calculated the total distance traveled as 27,500 meters. The total time taken is the sum of the time taken for each phase:
Total time = 10 min + 20 min + 5 min
Total time = 35 min
Converting the time to minutes:
Total time = 35 min * 60 s/min
Total time = 2100 s
Average velocity = 27,500 m / 2100 s
Average velocity ≈ 13.1 m/s
Therefore, the average velocity of the train is approximately 13.1 meters per second.