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The displacement  (x) of a particle is related to time t as x=a t+b t^2-c t^3, where a, b, c  are constants of motion. The velocity of the particle when its acceleration is zero is given by -

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

 x=a t+b t^2-c t^3 \\
v=\frac{d x}{d t}=a+2 b t-3 t^2 c-\text { (1) } \\
a=\frac{d^2 x}{d t^2}=0+2 b-6 c t

From question, acceleration is zero.

0=0+2 b-6 c t \\
t=\frac{b}{3 c}

Now, velocity will become-

v=a+\frac{2 b^2}{3 c}-\frac{3 b^2}{9 c^2} \cdot c \\
v=a+\frac{2 b^2}{3 c}-\frac{b^2}{3 c}=a+\frac{b^2}{3 c}
 

Posted by

Ajit Kumar Dubey

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