At the Aircraft Landing Dynamics Facility located atNASA\'s Langley Research Cen

Question

At the Aircraft Landing Dynamics Facility located atNASA's Langley Research Center in Virginia, a water-jet nozzlepropels a 46,000 kg sled from zero to 400 km/hour in 2.5seconds. (This sled is equipped with tires that are being testedfor the space shuttle program, which are then slammed into theground to see how they hold up.) (a) Assuming a constantacceleration for the sled, what distance does it travel during thespeeding-up phase? (b) What is the net work done on the sled duringthe time interval? Somewhere along the line I have got the distance to be = speed* time = .5 * (V1+V2) * time At the Aircraft Landing Dynamics Facility located atNASA's Langley Research Center in Virginia, a water-jet nozzlepropels a 46,000 kg sled from zero to 400 km/hour in 2.5seconds. (This sled is equipped with tires that are being testedfor the space shuttle program, which are then slammed into theground to see how they hold up.) (a) Assuming a constantacceleration for the sled, what distance does it travel during thespeeding-up phase? (b) What is the net work done on the sled duringthe time interval? Somewhere along the line I have got the distance to be = speed* time = .5 * (V1+V2) * time Somewhere along the line I have got the distance to be = speed* time = .5 * (V1+V2) * time

Explanation / Answer

   a.   By first equation ofmotion

         v   =   u+ a * t

   v   =   finalspeed   =    400km/h   =   400 * 5 /18   =   111.11   m/s

   u   =   initialspeed   =   0,   a   =   accelerationproduced,   t   =   time   =   2.5   s

      111.11   =   0+ a * 2.5

      a   =   111.11/2.5   =   44.44   m/s2

        Second equationof motion is      distancetrvalled         d   =   u* t + (1/2) * a * t2

    d   =   0* 2.5 + 0.5 * 44.44 *2.52   =   138.87   m

    b.   Workdone   =   Force (F) * distance(d)   

      W   =   (m* a) * d

         W   =   46000* 44.44 * 138.87   =   2.84 *108   J

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