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A train is targated to run from Delhi to Pune at an average speed of 80 kph but due to repairs of track looses 3 hrs in the first part of the journey. If then accelerates at a rate of 20 kph^2 till the speed reaches 100 kph. Its speed is now maintained till the end of the journey. If the train now reaches station in time, find the distance (in km) from when it has started acceleration.


A train is targated to run from Delhi to Pune at an average speed of 80 kph but due to repairs of track looses 3 hrs in the first part of the journey. If then accelerates at a rate of 20 kph^2 till the speed reaches 100 kph. Its speed is now maintained till the end of the journey. If the train now reaches station in time, find the distance (in km) from when it has started acceleration.


Grade:10

1 Answers

Ram Kushwah
110 Points
3 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
 
 
 

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