HCV: A train starts from rest and moves with a constant acceleration of 2.0m/s2 for half a minute.

[@RashiSinghDangi]

Got the real Vibe of hcv by this question 😶‍🌫️😵‍💫

[@AnkitKumar-y9l5z]

Very nice explanation

[@karthikeya3153]

Very nice explanation sir thanks for the video

[@amanmahajan5600]

better explanation than many out there

[@srishtidas-bg3xb]

Splendid explanation sir

[@bhaveshkumar1269]

Amazing sir

[@legendnayakyt4108]

Nice explanation

[@KrrishRoy-v3t]

Sir it would be better if u divide the question in 2 parts like For 30 sec and for 60 sec

[@smayra8694]

Ty sir❤️

[@KrrishRoy-v3t]

Maximum students are getting confused at acceleration = -1sec

[@srijamondal3003]

In the second case, where the train is decelerating, it's initial velocity was 60m/s and for applying brakes it decreased to 0. So, v is 0 u is 60m/s and ,t is 60seconds(given in the question) Thus by the first equation of motion,v=u+at(substituting the values) 0=60+a*60 a=-60/60=1m/s^2 Hope you got it

[@CuriousFacts468]

IIT Delhi 😮❤

[@AryanSingh-iy2tw]

sir last me 7:23 par 900+1350 hoga ya fir 225+1350 because question when the velocity is half?

[@shreyanshpatel9740]

1350+900 hoga ,bro

[@srijamondal3003]

The velocity will get halved at two places in the complete journey For the first time,when the train is accelerating,it's velocity will increase from 0m/s(rest) to 60m/s( max velocity).So in the mean time ,it's velocity at some point was 30m/s and we have to find the position where it was 30m/s for the first time,and it came out to be 225 metres. Then , the second case is occurring when the train is decelerating. The deceleration is found to be -1m/s^2.It means that the trains initial velocity in this case after applying the brakes is 60m/s ( v max),after which the train's velocity is decreasing to 0(rest).So here also in the mean time while decreasing from 60-0 ,in some point it's velocity was 30m/s. So we have to find this position where the velocity was 30m/s. We found it to be 1350 metres. This is the distance after reaching 60m/s already.But we have to also consider the previous 900 m distance as the exact position of the train is concerned here. Thus to get the position in second case where the train's velocity was 30m/s,we need to add the previous distance (900 m ) and the distance after reaching 60 m/s ,which is 1350 m.. which is coming out to be 2250 metres or 2.25 kms. Hope you understood, although it would be difficult for you to understand by a comment .It could have been better explained with the V vs T graph.

[@tanvigoyal2480]

Same here I also have the same doubt That if he is adding 900m with 1350 m then what is the use of 225m??🤔🤔

[@freefireking__1_]

U=60m/s kaise hua