The breaks applied to a train moving at 90km/h produce a retardation of 5m/s^2 what will be the distance it covers to a stop
Answers
Answered by
2
initial velocity=90kmph =25mps.use formula v²=u²-2as0=25²-2*5s625=10ss=62.5m
Answered by
0
the answer is 62.5 metre please mark me brainliest
Similar questions