Question

The relation between time t and distance x is t = $a{x}^2$+bx where a and b are constant. The acceleration is:

• A. $-2ab{v}^2$
• B. $2b{v}^2$
• C. $-2a{v}^2$
• D. $2a{v}^2$

Answer 1

The correct option is C $-2a{v}^2$

Reaction between ${}^{\prime }{x}^{\prime }$ and ${}^{\prime }{t}^{\prime }$ in given as:

$t=a{x}^2+bx$

difference $t$,

$\Rightarrow \frac{dt}{dx}=2ax+b$

so, velocity of a particles in $v=\frac{dx}{dt}$, from above we say:

$v=\frac{dx}{dt}=\frac{1}{dt/dx}=\frac{1}{2ax+b}\to \left(1\right)$ [using properties of derivatives]

similarly we can find acceleration:

$a_e=\frac{dv}{dt}=\frac{d}{dt}\left(\frac{1}{2ax+b}\right)$

$=\frac{-2a}{(2ax+b{)}^2}$ [using Quotient Rule for differentiation]

From $\left(1\right)$, we replace $\frac{1}{(2ax+b{)}^2}=by{v}^2$

so, $a_e=2a{v}^2$