Question

The motion of a particle is given by $t=\alpha {{x}^{2}}+\beta x$ where ‘t’ is time, ‘x’ is position, $'\alpha '$, $'\beta '$ are constants and ‘v’ will denote velocity. The acceleration at time ‘t’ is given by
(A) $-2\beta {{v}^{3}}$
(B) $-2\alpha {{v}^{3}}$
(C) $2\beta {{v}^{3}}$
(D) $2\alpha {{v}^{3}}$

Answer 1

Hint: We must have a brief idea about the Newton’s Laws of Motion and we must thoroughly remember the various equations relating to the motion of a particle, that is, its time period of motion, its velocity during motion, the distance it covers and the acceleration it achieves during its motion. Keeping this in mind, we can proceed with such questions.

Complete step-by step answer:
From Newton’s Laws of Motion, we can determine the equation of velocity of a moving particle. Accordingly, from the equation of velocity, we can determine the acceleration and time period of the particle in motion. From the given information, and the expression of t, we can evaluate the time period with simple differentiation.

$t=a x^{2}+b x,$ Differentiating with respect to time, we get

$\left( t \right)\dfrac{d}{dt}(t)=a\dfrac{d}{dt}\left( {{x}^{2}} \right)+b\dfrac{dx}{dt}=a\cdot

2x\dfrac{dx}{dt}+b\cdot \dfrac{dx}{dt}$

$1=2 a x v+b v=v(2 a x+b) \Rightarrow 2 a x+b=\dfrac{1}{v}$

Again, differentiating we get $2a\dfrac{dx}{dt}+0=\left( -\dfrac{1}{{{v}^{2}}}\dfrac{dv}{dt} \right)$

$\Rightarrow \dfrac{d v}{d t}=f=-2 a v^{3}$

Hence, the correct answer is Option 2.

Note: We must keep in mind the scenarios when we need to use differentiation, integration or double differentiation in case of such problems to overcome with the solutions. We must also be moderately familiar with various cases of differentiation including the rules of differentiation to be able to solve such questions.