a train starts from rest from station with acceleration 0.2 m/s2 on a straight track and then comes to the rest after attending maximim speed on another staition due to retardation 0.4 m/s2 if the total time spend is half hour then calculate the distance between two station
Answers
Answer:
Distance will be 216 km.
Explanation:
Since, from the question we will get that the train accelerates with acceleration of a = 0.2 m/s^2, and if we let t = t1 and v= v1.
So, applying newtons laws of motion v=u + at, since we know that the v1=0.2t1 or t1= 0.2 v1 /2 (1) .
Again, applying v2=v1+a2t2(when train is retarding and s =s1).
S=ut + 1/2at^2( where u will be zero).
s1=1/2 x 0.2 x (t1)^2 .
So, now v2=v1+a2t2.
0 = v1+0.4t2
t2=–v1/0.4 (2)
Now, applying them in the equation s2=v1t2 +(1 /2) x a2(t2)^2.
So, the total time will be t=t1+t2 = 30 min = 1800 seconds.
So, we get that t1=1200 secs and t2 = 600 secs.
So, then substituting the values of a, v and t in (1) and (3),
s1 = 144 km and s2= 72 km
Total distance will be s= s2+ s2.
s=216km
Answer: Distance between the two station is 216 km
Explanation:
Case I:
Train accelerates with a = 0.2 m/s2 , t = t1 and v= v1
v=u + at and v1=0.2t1 or t1= 0.2 v1 /2 …(1)
s=ut+(1/2)at2
Case II:
v2=v1+a2t2(train is retarding and s =s1)
s1=1/2 x 0.2 x t12
v2=v1+a2t2
0 = v1+0.4t2
t2=–v1/0.4 …(2)
s2=v1t2 +(1 /2) x a2t22
t=t1+t2 = 30 min = 1800
t1=1200 secs and t2 = 600 secs
substituting a, v and t in (1) and (3), we get
s1 = 144 km and s2= 72 km
Total distance s= s2+ s2
s=216km