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
Menu
Grade: 12th pass
        
A train starts from rest and moves with a constant acceleration of 2m/s for half a min .the brakes are then applied and the train comes to rest in one min after applying breaks. Find the total distance moved by train ,maximum speed attained bt the train,position of the train at half the max.speed
10 months ago

Answers : (2)

Khimraj
2729 Points
							
max speed = 2*30 = 60m/s
let retardation is a
then
60 – a*60 = 0
so a = 1m/s
time when train at half the max speed
30 = 2*t
t = 15 sec
position at time t = 15
= ½*2*152 = 225 m
total distance covered by train
d = ( ½*2*302 ) + (60*60 – ½*1*602)
d = (900) + (3600 – 1800) = 2700m
Hope it clears.
10 months ago
Rajdeep
227 Points
							
HELLO!
 
Train starts from rest, hence the initial velocity u = 0.
It moves with acceleration = 2m/s2 for half minute (30 seconds).
 
Distance covered in this time interval is given by:
S = ut + ½ at2
= 0 + ½ x 2 x 30 x 30
= 900m
 
Velocity attained by this acceleration after 30 seconds:
v = u + at
=> v = 0 + 2 x 30
=> v = 60 m/s
 
From this velocity, brakes are applied and train comes to rest in 60 seconds.
 
The retardation is given by:
 
v = u – at
=> 0 = 60 – a x 60
=> a = 1 m/s2
 
Distance covered in this time:
v2 = u2 + 2aS
=> 0 = (60)2 + 2 (-1) S
=> 0 = 3600 – 2S
=> S = 3600/2 = 1800m.
 
So, total distance moved = 900m + 1800m = 2700m.
Maximum speed of the train = 60 m/s.
 
Position of the train at half its maximum speed.
Here, you need to note that first the train is accelerating to 60 m/s, and then it is decelerating to 0 m/s. So there are two positions when speed is 30 m/s.
 
(I) When the train is accelerating with an acceleration of 2 m/s,
time at which speed = 30m/s is:
v = u + at
=> 30 = 0 + 2 x t
=> t = 15s
 
At 15s, distance covered from origin is:
S = ut + ½ at2
= 0 + ½ x 2 x 15 x 15
= 225m
 
(II) When the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 ,  time at which the speed reaches 30 m/s is:
v = u – at
=> 30 = 60 – 1xt
=> t = 30s
 
At 30s, distance covered is:
S = ut – ½ at2
= 60 x 30 – ½ x 1 x (30)2
= 1800 – (15 x 30)
= 1800 – 450
= 1350m (from the initial 900m covered).
 
So, distance from origin = 900 + 1350m = 2250m.
 
HENCE, THE TRAIN IS AT 225m and 2250m from its starting point, at half its maximum speed.
 
Thanks!
 
 
10 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details