Question

The relation between time $t$ and displacement $x$ is $t=\alpha x^2+\beta x$ where $\alpha$ and $\beta$ are constants then acceleration of particle is proportional to-

• A. $v^3$
• B. $v^2$
• C. $v^{3/2}$
• D. $v^{2/3}$

Answer 1

The correct option is A $v^3$

The relation between time $t$ and displacement $x$ is given by

$t=\alpha x^2+\beta x$. . .. . .. . . .(1)

Differentiate above equation with respect to $x$,

$\frac{dt}{dx}=2\alpha x+\beta$

and we know that

velocity, $v=\frac{dx}{dt}$

$v=\frac{1}{2\alpha x+\beta }$. . . . . .(2)

acceleration, $a=\frac{dv}{dt}$

$a=-\frac{1}{(2\alpha x+\beta {)}^2}2\alpha \frac{dx}{dt}$

$a=-2\alpha v^3$ (by putting equation 2)

The acceleration of the particle is directly proportional to the cubic of velocity.

$a\propto v^3$

The correct option is A.