Question

The relation between the time t and position x for a particle moving on the x-axis is given by $t=p{x}^2+qx$, Where $p$ and $q$are constants. The relation between velocity $b$and acceleration $a$ is as

• A. $a\alpha v^3$
• B. $a\alpha v^2$
• C. $a\alpha v^4$
• D. $a\alpha v^{}$

Answer 1

The correct option is A $a\alpha v^3$

The relation between the time$t$ and position $x$ for a particle moving on the x-axis is given by

$t=p{x}^2+gx$

Differentiate with respect to time,

$\frac{dt}{dt}=\frac{d\left(p{x}^2\right)}{dt}+\frac{d\left(qx\right)}{dt}$

$1=2px\frac{dx}{dt}+q\frac{dx}{dt}$. . . .. .(1)

We know that, velocity $v=\frac{dx}{dt}$

Equation (1) becomes

$1=2pxv+qv$

$x=\frac{1-qv}{2pv}$. . . . . .(2)

Differentiate with respect to time again

$\frac{d\left(1\right)}{dt}=2p\frac{d\left(xv\right)}{dt}+q\frac{dv}{dt}$

$0=2p(x\frac{dv}{dt}+v\frac{dx}{dt})+q\frac{dv}{dt}$

$2p(xa+v.v)+qa=0$ ($a=\frac{dv}{dt}$)

$2pxa+2p{v}^2+qa=0$

$2p\frac{1-qv}{2pv}a+2p{v}^2+qa=0$ (from equation 2)

$\left(\frac{1-qv}{v}\right)a+qa=-2p{v}^2$

$a(1-qv+qv)=-2p{v}^3$

$a=-2p{v}^3$

$a\propto v^3$

The correct option is A.