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Q.

If a train travelling at 72 kmph is to be brought to rest in a distance of 200 metres, then its retardation should be

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a

20  ms-2

b

10 ms-2

c

2 ms-2

d

1 ms-2

answer is D.

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Detailed Solution

u=72kmph=20m/s,v=0By using v2=u2−2as⇒a=u22s=(20)22×200=1m/s2
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