Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

a train starts from rest from a station with accleration 0.2 on a staight track and then comes to rest after attaning maximum speed on another station due to retardation 0.4 ..if total time spent is half an hour, then distance between two stations is {neglect length of train}...

a train starts from rest from a station with accleration 0.2 on a staight track and then comes to rest after attaning maximum speed on another station due to retardation 0.4 ..if total time spent is half an hour, then distance between two stations is {neglect length of train}...

Question Image
Grade:12th pass

1 Answers

Arun
25763 Points
3 years ago
Dear Shubham
 
Case I:

Train accelerates with a = 0.2 m/s, t = t1and v= v1

v=u + at and v1=0.2tor 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

 

Regards

Arun (askIITians forum expert)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free