30-09-2013, 05:48 PM
You are absolutely right. I see that I wrote:
The deceleration is 0.5m/s/s which means that it take 1 second for the speed to be reduced by 1 m/s. Therefore with initial speed of basically 39m/s then it takes 39s to stop!
Let's change that to read:
The deceleration is 0.5m/s/s which means that it take 2 seconds for the speed to be reduced by 1 m/s. Therefore with initial speed of basically 39m/s then it takes 78s to stop. Somehow I followed your incorrect lead- to be honest whilst replying to your other post yesterday the thought did go through my mind that this one was wrong but I then failed to go back and check so you beat me to it.
The deceleration is 0.5m/s/s which means that it take 1 second for the speed to be reduced by 1 m/s. Therefore with initial speed of basically 39m/s then it takes 39s to stop!
Let's change that to read:
The deceleration is 0.5m/s/s which means that it take 2 seconds for the speed to be reduced by 1 m/s. Therefore with initial speed of basically 39m/s then it takes 78s to stop. Somehow I followed your incorrect lead- to be honest whilst replying to your other post yesterday the thought did go through my mind that this one was wrong but I then failed to go back and check so you beat me to it.
(30-09-2013, 08:10 AM)prpaoorna Wrote: Hi PJW/Peter,
I found a mistake in stopping calculations.
Deceleration time is supposed to be 38.89/a=78sec. hence
deceleration distance=78*19.45=1517m (because avg.speed is 19.45m/s), but calculation shows deceleration time as 19.45/a=39sec. so deceleration distance was the 759m instead of 1517m
what is the correct approach here?
(24-09-2013, 06:14 AM)prpaoorna Wrote: Hi all,
could you please check my calculations?
Line to be covered in between pages is “Non-stopping train will cover the 1698m in 44 sec.”
Regards,
N Prapoorna.
PJW