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9th Class Science Time and Motion Question Bank Motion

  • question_answer
    A particle is moving in a straight line with initial velocity u and uniform acceleration a. If the sum of the distances travelled in \[{{t}^{th}}\] and\[{{(t+1)}^{th}}\]seconds is 100 cm, then its velocity after \[t\] seconds in \[cm\text{ }{{s}^{1}}\] is

    A) 20                               

    B) 30                   

    C) 50                   

    D) 80

    Correct Answer: C

    Solution :

    Distance travelled in \[{{t}^{th}}\] second of uniformly accelerated motion is \[{{s}_{t}}=u+\frac{a}{2}(2t-1)\] - (i) Distance travelled in \[{{(t+1)}^{th}}\] second can be written as \[{{s}_{t+1}}=u+\frac{a}{2}[2(t+1)-1]\] Or \[{{s}_{t+1}}=u+\frac{a}{2}(2t+1)\] \[\because {{s}_{1}}+{{s}_{t+1}}=100cm\] (given) \[\therefore u+\frac{a}{2}(2t-1)+u+\frac{a}{2}(2t+1)=100\] Or \[u+at-\frac{a}{2}+u+at+\frac{a}{2}=100\] Or \[2u+2at=100\] or \[u+at=50\] \[\therefore v=50\,\,cm\,\,{{s}^{-1}}\]


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