Question

The relation between time $t$ and displacement $x$ is $t=\alpha x^2+\beta x$ where $\alpha$ and $\beta$ are constants. The retardation is

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

Answer 1

The correct option is A $2\alpha v^3$

Differentiate the given relation with respect to time to get,

$1=2\alpha x\frac{dx}{dt}+\beta \frac{dx}{dt}$
or, $1=2\alpha xv+\beta v$

where $\frac{dx}{dt}=v$
Thus we have,

$2\alpha x+\beta =\frac{1}{v}...................\left(1\right)$
Differentiating again and using product rule $\frac{d}{dt}\left(uv\right)=u\frac{dv}{dt}+v\frac{du}{dt}$ we get,

$0=2\alpha x\frac{dv}{dt}+2\alpha v\frac{dx}{dt}+\beta \frac{dv}{dt}$
or, $0=a(2\alpha x+\beta )+2\alpha v^2$

where $\frac{dv}{dt}=a$
Use $\left(1\right)$ to get, $0=\frac{a}{v}+2\alpha v^2$ or,

$Retardation=-a=2\alpha v^3$