Question

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

Answer 1

The correct option is A $-2a{v}^3$
$t=a{x}^2+bx$

Differentiate w.r.t time

$1=2ax\frac{dx}{dt}+b\frac{dx}{dt}$

$\frac{dx}{dt}=\frac{1}{2ax+b}$

$V=\frac{1}{2ax+b}$...(1)

$a=\frac{dV}{dt}=\frac{-\left(2a\frac{dx}{dt}\right)}{(2ax+b{)}^2}$

$a=\frac{-2av}{(2ax+b{)}^2}$....(2)

Substiting $\frac{1}{2ax+b}=V$ in (2)

we will get, $a=-2a{v}^3$