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a train travels from one station to the next. The drover of train A starts from time t=0 and accelerates uniformly for the first 20 s. At time t=20 s, train reaches its top speed of 25m/s , then travels at this speed for further 30 s before decelerating unoformly to rest. Total time for the journey of train A is 60 s. Calculate the deceleration of train A , as it comes to rest.

I went through the explanation but why have they taken the time as 10 s when its 20 seconds ????

It is not clear which explanation you are referring to. Please find below the answer to the above question which you can match with the explanation with you. This may clear your doubt.

Given,

- Initial velocity, u = 0 m/s
- Initial time, t = 0 s
- Velocity reached in t = 20 s, v = 25 m/s

Then

from the kinematic relation,

$v=u+at\phantom{\rule{0ex}{0ex}}v=at\phantom{\rule{0ex}{0ex}}a=\frac{v}{t}=\frac{25}{20}=1.25m/{s}^{2}$

The train moves at v = 25 m/s for 30 s.

Then the train decelerates uniformly to rest.

Total journey time = 60 s

Time within which the velocity, v = 25 m/s reduces to v' = 0 m/s is t = 60 - (20+30) = 10 s

So, using the kinematic relation,

$v\text{'}=v+a\text{'}t\phantom{\rule{0ex}{0ex}}0=25+10a\text{'}\phantom{\rule{0ex}{0ex}}a\text{'}=-2.5m/{s}^{2}$

So, the train decerates at a rate 2.5 m/s

^{2}.

Regards

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