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  2. Solve! - Kinematics - JEE Main-2
# Engineering

The position of a particles as a function of time t , is given by

x(t)=at+bt^{2}-ct^{3}

where a,b,and c are constants . When the particle attains zero acceleration , then its velocity will be :

  • Option 1)

    a+\frac{b^{2}}{4c}

  • Option 2)

    a+\frac{b^{2}}{3c}

  • Option 3)

    a+\frac{b^{2}}{c}

  • Option 4)

    a+\frac{b^{2}}{2c}

 

Answers (1)
S solutionqc

Given 

x(t)=at+bt^{2}-ct^{3}

v=\frac{dx}{dt}=a+2bt-3ct^{2}

a=\frac{dv}{dt}=2b-6ct

a=0 \Rightarrow t =\frac{2b}{6c}= \frac{b}{3c}

V \left ( t=\frac{b}{3c} \right )=a+2b\times \frac{b}{3c}-3c\times \frac{b^{2}}{9c^{2}}

V=a+\frac{b^{2}}{3c}


Option 1)

a+\frac{b^{2}}{4c}

Option 2)

a+\frac{b^{2}}{3c}

Option 3)

a+\frac{b^{2}}{c}

Option 4)

a+\frac{b^{2}}{2c}

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