The position x of a particle moving in a straight line varies with tim

1.	The position x of a particle moving in a straight line varies with time t as x=(t+3)^-1. The acceleration of particle is proportional to
Answer» Answer: Option (1) $(\text{velocity})^{3/2}$ Explanation: GIVEN, the position x VARIES with time t as $x=(t+3)^{-1}$ or, $x=\frac{1}{x+3}$ We KNOW that rate of CHANGE of displacement is velocity i.e. $v=\frac{dx}{dt}$ or, $v=\frac{d}{dt}(t+3)^{-1}$ or, $v=-(t+3)^{-2}$ Also, the rate of change of velocity is accelration i.e. $a=\frac{dv}{dt}$ or, $a=\frac{d}{dt}[-(t+3)^{-2}]$ $\implies a=2(t+3)^{-3}$ $\implies a=2[(t+3)^{-2}]^{3/2}$ $\implies a=2[-v]^{3/2}$ $\implies a=-2v^{3/2}$ Thus, $a \propto v^{3/2}$ or, $\text{acceleration}\propto (\text{velocity})^{3/2}$