Physics
posted by Dan on .
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.

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.