Physics

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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 -

    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=11.2-v/120(11/4+6/4)

    time constant speed= 41/4-11/4-3/2 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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