Question

A particle travels according to the equation $t=\alpha x^2+\beta x+\gamma$. The acceleration at any instant is

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

Answer 1

The correct option is B $-2{\alpha v}^3$

$t=\alpha x^2+\beta x+\gamma$
Differentiate with respect to time,

$1=2\alpha x\frac{dx}{dt}+\beta \frac{dx}{dt}or,1=2\alpha x.v+\beta .v.........................\left(1\right)$
Again differentiate with respect to time,

$0=2\alpha x\frac{dv}{dt}+2\alpha v\frac{dx}{dt}+\beta \frac{dv}{dt}$
or, $0=2\alpha xa+2\alpha v.v+\beta a$

$Thus,a=\frac{-2\alpha v^2}{2\alpha x+\beta }$ .
Using $\left(1\right),$ we have, $a=-2\alpha v^3$