Question

equations

Answer 1

@midhun.madhu1987 sorry to keep bothering you but am i doing this right?

Answer 2

OMG!!! is it question or essay?

Answer 3

its a question

Answer 4

i don't think we need 15.0 because thats a trick question @midhun.madhu1987

Answer 5

@armi

Answer 6

2.0 m/s^2 is the deceleration?

Answer 7

We have to consider 22' even for the first and last station?

Answer 8

oh okay

Answer 9

22'' sorry xD

Answer 10

is it a yes for both? :)

Answer 11

yes, I mean does the train start from station one at t=0 or t=22? and it ends when it reaches the last station and it waits 22'' ? or when it just arrives?

Answer 12

the tainst arts from 0 and waits a 22s

Answer 13

Ok the three equations u need are
\[v=\delta x/\delta t\]
for linear motion v=k
and for a=k
\[x=a/2t^2\]
where x is the space distanced
and \[v=at+v_o\]
which gives basically how much time you need to accelerate to get that speed.
Now, to go from Station A to B you need to consider:

The first 22'', the time it gets to reach that speed (3rd formula), the time (this after because you don't know the space yet) in which it drive at constant speed, finally the time to decelerate (always 3rd formula).

Multiply everything by 6 and u get it. My outcome is roughtly 15 minutes and 12s

To calculate the time at constant speed u need space which is (3000-the space it covers while accelerating-the space it covers while decelerating)
For these u use equation number 2.
Any doubt or misunderstanding just ask ;)

Answer 14

i'll see what i come up with...