Posted by Dan on Sunday, July 11, 2010 at 12:34pm.
It took a stopping train 41/4min to travel between two stations 11.2km apart.The train accelerated uniformly during the first 11/4min of the journey,and decelerated uniformly to rest during the last 3/2min of the journey.The train had moved with uniform speed during the remaining part of its travel.calculate (a)Uniform speed,km/h (b)Uniform acceleration (c)Uniform retardation (d)Distance travelled in the first 4min of the journey (c)Distance covered in the last 3min of the journey.

Physics  bobpursley, Sunday, July 11, 2010 at 2:48pm
assume the maximum velocity was V.
Then the average velocity during starting and stopping was V/2.
So, during starting, you know the avg velocity, and time, so
distancestarting=v/2 * 11/4*1/60 in km if v is in km/hr
distancestopping= v/2*3/2*1/60
distance constant speed=11.2v/120(11/4+6/4)
time constant speed= 41/411/43/2 min
convert that to hours.
velocity v= distanceconstantspeed/timeconstant speed
from that calculate v.
Everything else ought to work out easily.
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