Asked by emma
with an initial speed of 260km/h a train takes 1.50 x 10^3 m to stop on a level track. assuming the 'deceleration' is constant. what is the time in seconds taken for stopping?
Answers
Answered by
Damon
260,000 m/h * 1 h /3600 s = 72.2 m/s = Vinitial
average speed during deceleration = 72.2 / 2
= 36.1 m/s
goes 1500 m at 36.1 m/s
1500/36.1 = 41.5 seconds
average speed during deceleration = 72.2 / 2
= 36.1 m/s
goes 1500 m at 36.1 m/s
1500/36.1 = 41.5 seconds
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