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A particle maviong alonge a stright line travels one third of the total distance with a speed of 3 m/s the remaining distance is covered with a speed of 4m/s for half the time and 5m/s for other half of the time .The average speed during the motion is ?
Asked by RAHUL rd | 01 Nov, 2014, 04:19: PM
answered-by-expert Expert Answer
Let S be the total distance travelled by the particle.
begin mathsize 14px style bold Given bold space bold at bold space bold distance bold space bold equals bold space bold S over bold 3 comma space the space speed space of space the space particle space equals space 3 space straight m divided by straight s We space know space that space Speed space equals fraction numerator Distance space over denominator time end fraction rightwards double arrow Time space equals Distance over Speed equals fraction numerator straight S over denominator 3 cross times 3 end fraction equals straight S over 9 Let space this space time space taken space by space the space particle space for space the space first space one space third space distance space be space straight t subscript 1 rightwards double arrow straight t subscript 1 space equals straight S over 9 Now space the space remaining space distance space equals space fraction numerator 2 straight S over denominator 3 end fraction space. Let space straight t subscript 2 space be space the space time space taken space for space the space remaining space half. The space particle space travelled space with space straight a space speed space of space 4 straight m divided by straight s space in space first space fraction numerator space straight t subscript 2 over denominator 2 end fraction and space then space with space straight a space speed space of space 5 space straight m divided by straight s space in space next space fraction numerator space straight t subscript 2 over denominator 2 end fraction We space know space that space distance space equals space speed cross times time Hence space we space can space write space that space the space remaining space distance space fraction numerator 2 straight S over denominator 3 end fraction space equals 4 cross times fraction numerator space straight t subscript 2 over denominator 2 end fraction plus 5 cross times fraction numerator space straight t subscript 2 over denominator 2 end fraction straight i. straight e. space fraction numerator 2 straight S over denominator 3 end fraction equals fraction numerator 4 straight t subscript 2 over denominator 2 end fraction plus fraction numerator 5 straight t subscript 2 over denominator 2 end fraction fraction numerator 2 straight S over denominator 3 end fraction equals fraction numerator 9 straight t subscript 2 over denominator 2 end fraction straight t subscript 2 equals fraction numerator 4 straight S over denominator 27 end fraction So space thus space the space total space time space straight T space equals straight t subscript 1 plus straight t subscript 2 straight T equals straight S over 9 plus fraction numerator 4 straight S over denominator 27 end fraction space space equals fraction numerator 7 straight S over denominator 27 end fraction Total space distance space equals straight S Therefore space average space speed space equals fraction numerator Total space distance over denominator total space time end fraction equals fraction numerator straight S over denominator begin display style fraction numerator 7 straight S over denominator 27 end fraction end style end fraction equals fraction numerator 27 straight S over denominator 7 straight S end fraction equals 3.85 space straight m divided by straight s end style
Thus the average speed during the motion = 3.85 m/s
Answered by Jyothi Nair | 02 Nov, 2014, 11:01: AM

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