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 can break it down into three different phases: acceleration, constant velocity, and deceleration.
Step 1: Acceleration phase
Given:
Initial velocity (u) = 0 m/min
Acceleration (a) = 100 m/min^2
Time (t) = 10 min
We can use the constant acceleration formula to find the maximum velocity reached during this phase:
v = u + at
Substituting the values:
v = 0 + (100 m/min^2) * (10 min)
v = 1000 m/min
So, the maximum velocity reached during the acceleration phase is 1000 m/min.
Step 2: Constant velocity phase
During this phase, the velocity remains constant. We are given that this phase lasts for 20 minutes. Therefore, the velocity during this phase will be the same as the maximum velocity reached during the acceleration phase, which is 1000 m/min.
Step 3: Deceleration phase
Given:
Initial velocity (u) = 1000 m/min (the velocity at the end of the constant velocity phase)
Time (t) = 5 min (the time taken to come to rest)
We need to find the retardation during this phase. Retardation is the negative acceleration. Thus, we can use the constant acceleration formula to find the retardation:
v = u + at
Substituting the values:
0 = 1000 m/min + a * (5 min)
Solving for a:
-1000 m/min = 5a
a = -200 m/min^2
So, the retardation during the last 5 minutes is -200 m/min^2.
Step 4: Total distance traveled
To find the total distance traveled, we need to calculate the distance covered during each phase separately and then sum them up.
Distance covered during acceleration phase:
s1 = ut + 0.5at^2
Substituting the values:
s1 = 0 * (10 min) + 0.5 * (100 m/min^2) * (10 min)^2
s1 = 0 + 5000 m
s1 = 5000 m
Distance covered during constant velocity phase:
s2 = v * t
Substituting the values:
s2 = (1000 m/min) * (20 min)
s2 = 20000 m
Distance covered during deceleration phase:
s3 = ut + 0.5at^2
Substituting the values:
s3 = (1000 m/min) * (5 min) + 0.5 * (-200 m/min^2) * (5 min)^2
s3 = 5000 m + (-2500 m)
s3 = 2500 m
Total distance traveled:
Total distance = s1 + s2 + s3
Total distance = 5000 m + 20000 m + 2500 m
Total distance = 27500 m
So, the total distance traveled by the train is 27500 meters.
Step 5: Average velocity
Average velocity is given by the total distance traveled divided by the total time taken.
Total time taken = 10 min + 20 min + 5 min
Total time taken = 35 min
Average velocity = Total distance / Total time taken
Average velocity = 27500 m / 35 min
Average velocity = 785.71 m/min
So, the average velocity of the train is approximately 785.71 m/min.