Question

A train starts from rest moves with a constant acceleration of 2m/s^2 for half a minute. The brakes are then applied and then train comes to rest in one minute after applying breaks.
(c) the position(s) of the train at half the maximum speed

Answer 1

O_o
I thought i was tagged !

Answer 2

Yes u were

Answer 3

Same process !

Answer 4

I am not getting the correct answer

Answer 5

Max speed = 60

Answer 6

that is correct

Answer 7

yes

Answer 8

Now how to proceed

Answer 9

Use (30)\(^2\) = 0 + 2*2*s

Answer 10

Find s

Answer 11

YEs man got it!

Answer 12

why can't we use

s= ut +1/2 at^2

Answer 13

u can but first u will have to calculate t

Answer 14

by using V=U+at

Answer 15

ok got it

Answer 16

thanks!!!!

Answer 17

so it's better to use V^2=U^2+2as

Answer 18

Don't you think openstudy , wastes much time

Answer 19

how ?

Answer 20

I am doing this problem for last 35 minutes or so , and it was rather a simple one , and i have thousands yet to do

Answer 21

This medium is not useful for bigger problems

Answer 22

unless u have a lot of time

Answer 23

what will be the accelration at the instant when it is at maximum speed

Answer 24

That's true somehow....there are some input limitations...!
but if u have a touchscreen then u can do it faster by using drawing tool

Answer 25

Still , one problem i posted other day took 2.5 hours

Answer 26

and i didn't realise

Answer 27

Time flies away when u r on OS

Answer 28

I'll suggest its better to solve by urself..post them on os at extreme cases. :D

Answer 29

btw inst accn at the time of max speed will be 2m/s^2

Answer 30

is it , i framed this question

Answer 31

Since the train moves with const accn

Answer 32

yeah it will 2m/s^2

Answer 33

let's check by equation

Answer 34

v=u+at
60 =(30)a
yes 2 hehe

Answer 35

sorry for trouble

Answer 36

LOL !
It's ok :)

Answer 37

btw u saw the webpage i developed ?

Answer 38

nopess

Answer 39

http://openstudy.com/updates/53c51e71e4b00f624a92dfc1