JEE Main & Advanced Physics Motion in a Straight Line / सरल रेखा में गति Question Bank Non-uniform Motion

  • question_answer
    The position \[x\] of a particle varies with time \[t\] as \[x=a{{t}^{2}}-b{{t}^{3}}\]. The acceleration of the particle will be zero at time \[t\] equal to                      [CBSE PMT 1997; BHU 1999; DPMT 2000; KCET 2000]

    A)             \[\frac{a}{b}\]

    B)             \[7.5\ km/h\]

    C)             \[\frac{a}{3b}\]          

    D)             Zero

    Correct Answer: C

    Solution :

                    \[\frac{dx}{dt}=2at-3b{{t}^{2}}\Rightarrow \frac{{{d}^{2}}x}{d{{t}^{2}}}=2a-6bt=0\Rightarrow t=\frac{a}{3b}\]


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