question_answer

What is the retardation of a moving particle if the relation between time and position is \[\text{t}=\text{A}{{\text{x}}^{\text{3}}}+\text{ B}{{\text{x}}^{\text{2}}}\], where A and B are appropriate constants.

Answer:

                As, \[\text{t}=\text{A}{{\text{x}}^{\text{3}}}+\text{ B}{{\text{x}}^{\text{2}}}\] Differentiating it w.r.t. time, we have \[\text{1}=\left( \text{3A}{{\text{x}}^{\text{2}}}+\text{2Bx} \right)\text{dx}/\text{dt}\] \[\therefore \] velocity, \[\text{v}=\frac{dx}{dt}={{\left( \text{3 A}{{\text{x}}^{\text{2}}}+\text{2Bx} \right)}^{-\text{1}}}\].......(i) Acceleration, \[\frac{dv}{dt}=\left( -\text{1} \right){{\left( \text{3 A}{{\text{x}}^{\text{2}}}+\text{ 2Bx} \right)}^{-\text{2}}}\times \left( \text{6Ax }+\text{2B} \right)\frac{dx}{dt}\] \[\therefore \]\[\frac{-(6Ax+2B)}{{{(3A{{x}^{2}}+2Bx)}^{2}}}\times {{\left( \text{3A}{{\text{x}}^{\text{2}}}+\text{2Dx} \right)}^{-\text{1}}}\] Retardation, =\[~-\text{a}=\frac{6Ax+2B}{{{(3A{{x}^{2}}+2Bx)}^{3}}}\]