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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Application in Mechanics and Rate Measurer

  • question_answer
    The equation of motion of a car is \[s={{t}^{2}}-2t\], where t is measured in hours and s in kilometers. When the distance travelled by the car is \[15\,km\], the velocity of the car is

    A)            \[2\,km/h\]

    B)            \[4\,km/h\]

    C)            \[2\,km/h\]

    D)            \[8\,km/h\]

    Correct Answer: D

    Solution :

               Given equation \[s={{t}^{2}}-2t\]                  .....(i)            and \[\frac{ds}{dt}=\]velocity \[\frac{dS}{dt}=8\pi r\frac{dr}{dt}=\frac{8\pi .7.5}{28\pi }=10\,c{{m}^{2}}/\min \]                   .....(ii)            Put \[s=15\] in (i), we get \[{{t}^{2}}-2t-15=0\]                                         or \[(t-5)(t+3)=0\]            \[\therefore t=5\]            Hence velocity of car \[=v=2(5)-2=8\,\,km/h\].


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