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.
a. V = Vo + at
V = 0 + 100m/min^2 * 10min = 1000 m/min
b. a=(V-Vo)/t=(0-1000) / 5=-200m/min^2
c. d1 = 0.5a*t^2 = 50*10^2 = 5000 m.
d2 = 1000m/min * 20min = 20,000 m.
d3 = Vo*t + 0.5a*t^2
d3 = 1000*5 + (-100)*5^2 = 2500 m.
d1+d2+d3 = 5000 + 20000 + 2500=27,500 m
= Tot. dist. traveled.
d. T = t1 + t2 + t3 = 10+20+5 = 35 Min.
Va=(d1+d2+d3)/T=27,500m/35min=786 m/min
= Average velocity.
To find the maximum velocity reached by the train, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity (which is 0 since the train starts from rest), a is the acceleration, and t is the time.
In this case, the train accelerates uniformly at 100m/min^2 for 10 minutes. Converting 10 minutes to seconds, we have 10 * 60 = 600 seconds.
Using the equation, we can calculate the maximum velocity:
v = 0 + (100m/min^2) * 600s
v = 0 + 60000m/min
v = 60000 m/min
Therefore, the maximum velocity reached by the train is 60000 m/min.
To find the retardation in the last 5 minutes, we can use the same equation as above, but with negative acceleration since the train is decelerating.
In this case, the train uniformly retards and comes to rest in the next 5 minutes. Converting 5 minutes to seconds, we have 5 * 60 = 300 seconds.
Using the equation, we can calculate the retardation:
0 = v + (-a) * 300s
-300a = v
a = -v/300
Since the train comes to rest (v = 0), the retardation can be calculated as:
a = -0/300
a = 0 m/min^2
Therefore, the retardation in the last 5 minutes is 0 m/min^2.
To find the total distance traveled by the train, we need to calculate the distances covered during each phase of motion and then add them together.
During the first phase, the train accelerates uniformly for 10 minutes. Since we know the acceleration (100m/min^2) and the time (10 minutes), we can use the equation:
s = ut + (1/2)at^2
where s is the distance traveled, u is the initial velocity (0 for the train starting from rest), a is the acceleration, and t is the time.
Using this equation, we can calculate the distance covered during the first phase:
s1 = (0) * 10 + (1/2) * (100m/min^2) * (10^2 min^2)
s1 = 5000 m
During the second phase, the train maintains a constant velocity for 20 minutes. Since the velocity is constant, we can simply multiply the velocity by the time to get the distance:
s2 = (60000 m/min) * 20 min
s2 = 1200000 m
During the third phase, the train decelerates uniformly for 5 minutes until it comes to rest. Similar to the calculation in the second phase, we can multiply the velocity by the time:
s3 = (0 m/min) * 5 min
s3 = 0 m
Adding all the distances together, we get the total distance traveled:
total distance = s1 + s2 + s3
total distance = 5000 m + 1200000 m + 0 m
total distance = 1205000 m
Therefore, the total distance traveled by the train is 1205000 m.
To find the average velocity of the train, we can use the equation:
average velocity = total distance / total time
The total time can be calculated as the sum of the times for each phase of motion:
total time = 10 min + 20 min + 5 min
total time = 35 min
Converting 35 minutes to seconds, we have 35 * 60 = 2100 seconds.
Using the equation, we can calculate the average velocity:
average velocity = 1205000 m / 2100 s
average velocity = 574.7 m/s
Therefore, the average velocity of the train is approximately 574.7 m/s.