Question

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

Answer: (A)

$-2\alpha v^{-3}$

t$=\alpha x^2+\beta x$

Differentiating with respect to x we get

$\frac{dt}{dx}=2\alpha x+\beta v=\frac{1}{2\alpha x+\beta }\frac{dv}{dt}=a=-2\alpha (2\alpha x+\beta {)}^{-3}=-2\alpha v^{-3}$