#### Thank you for registering.

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

Please check your email for login details.

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

# ( IMP QUESTION). If the greatest admissible acceleration or retardation of a train be 3 feet/sec2.find the least time from one station to another at a distance of 10 metre the maximum speed being 60 miles per hour:(A) 500 sec (b) 58.67 sec (c) 400 sec this question is from arhiant bitsat 2014 book from motion chapter question no. Is 21 and the ans of question given is B but not solutionof problem. ?

Grade:12th Pass

## 2 Answers

Nirmal Singh.
askIITians Faculty 44 Points
7 years ago

a = 3 feet / sec^2
since 1 feet = 0.3048 m
so a = 3 *0.3048 = 0.9144 m/sec^2
s = 10 m
u = 60 miles/hr
1 miles = 1.61 km
so u = 60*1.61 km/hr = 60*1.61*5/18 km/hr
u = 26.8 m/sec
hence s = ut -1/2at^2
10 = 26.8t - 0.5*0.9144t^2
so 0.4592t^2 - 26.8t+10 = 0
t = 26.8+sqrt(26.8^2 - 4*0.4592*10) / 2*0.4592
t = 58.67 sec

Regards,
Nirmal SIngh
Askiitians Faculty

Sachin Bhutani
40 Points
7 years ago
thanks

## ASK QUESTION

Get your questions answered by the expert for free