#### Thank you for registering.

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

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
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

# The greatest acceleration or deceleration that a train may have is a .The minimum time in which the train may reach feom one station to the other seperated by a distance d isAns-(4d/a)1/2

Bhanu Pratap
25 Points
2 years ago
The total distnace train has to cover is d.
The first half it will accelerate and the second half it’ll deaccelerate, and it’ll take half-half time for both acceleration and deacceleration.

Hence,

d/2 = a/2(t/2)2               (u = 0 ,because the train was at rest initially)
4d/a = t2
t = (4da)½.

;)

Kaushki
19 Points
2 years ago
Sorry,but here it is not given that the train accelerate in half of total time.......................................?..........
Bhanu Pratap
25 Points
2 years ago
It is given that the train will either accelerate or deaccelerate i.e train cannot be in an uniform motion. the magnitude of acceleration and deacceleration is also same i.e a. Hence after accelerating for some time,the train will need the same amount of time to stop at the given retardation rate. u can”t increase or decrease the retardation rate, and hence u cannot decrease the time it”ll take to stop.
Aman Gupta
24 Points
2 years ago
But according to question we have to find the minimum time. At time will be minimum when the acceleration is high..... So we can't assume that it will decelerate half the way... If it will decelerate then the time would not be minimum???......