Ram Kushwah
Last Activity: 4 Years ago
Let the train was planned to reach Pune in T hrs
So toal distance to cover=80T km
Train is travelling in 3 steps
Sytep I : Delhi to point of acceleration
Let in first step train travells x1 distance
Here 3 hrs are lost
Let time in this step= t1 hrs
then x1=(t1-3)*80 Km...............................(1)
Time left =T-t1
Step II : Train accelerates from 80 km/hr to 100 km/hr
Here u=80 km/hr,a=20km/hr², v=100 km/hr
v=u+at
100=80+20*t
20t=20,t =1 hr
So t2=1 hr
Distance covered in 1 hr
=S=ut+1/2at²
=80*1+1/2* 20*1
=80+10=90 km
so x2=90 km...............................(2)
Time left= T-t1-1
STEP III : TRAIN WITH SPEED OF 100 Km/hr REACHES PUNE:
Distance covered
x3=(T-t1-1)*100).....................................(3)
Now x1+x2+x3=80T
Putting values of x1,x2 and x3 we get
(t1-3)*80 +90+(T-t1-1)*100=80T
80t1-240+90+100T-100t1-100=80T
-20t1-250+100T=80T
20T=250+20t1
2T=25+2t1
2(T-t1)=25
T-t1=12.5
t1=T-12.5
Now distance covered before acceleration
=x1=80(t1-3) From(1)
=80*(T-12.5-3)
=80T-80*15.5
=80T-1240
x1= Distance from Delhi to Pune – 1240
Thus train start accelerating at a distance of 1240 km from PUne
