Question

For a particle moving along x axis , time is given by t = ax^2 + ß x , where a and ß are constants . Find the retardation of the particle ? ​

Answer 1

Answer:

$\sf -\alpha v^3$

Explanation:

According to Question ,

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

Differenciate both sides wrt t .

$\sf\longrightarrow \dfrac{dt}{dt}= \dfrac{d}{dx} ( \alpha x^2 +\beta x) \\\\\\\sf\longrightarrow 1 = \dfrac{d}{dt}( \alpha x^2) +\dfrac{d}{dx}( \beta x ) \\\\\\\sf\longrightarrow 1 = \alpha 2x . \dfrac{dx}{dt} +\beta \dfrac{dx}{dt} \\\\\\\sf\longrightarrow 1 = 2\alpha x v +\beta v \\\\\\\sf\longrightarrow 1 = v ( 2\alpha x + \beta ) \\\\\\\sf\longrightarrow \boxed{\sf Velocity = \dfrac{1}{2\alpha x +\beta } }$

We had ,

$\sf\longrightarrow 1 = 2\alpha x v +\beta v \\\\\\\sf\longrightarrow \dfrac{d(1)}{dt}= \dfrac{ d (2\alpha x v +\beta v)}{dt} \\\\\\\sf\longrightarrow 0 = 2\alpha \dfrac{ d(xv)}{dt} +\beta \dfrac{dv}{dt} \\\\\\\sf\longrightarrow 0 = 2\alpha \bigg( \dfrac{dv}{dt} .x + v .\dfrac{dx}{dt} \bigg) +\beta x \\\\\\\sf\longrightarrow 0 = 2\alpha a x + 2\alpha v^2 +\beta a \\\\\\\sf\longrightarrow 0 = a ( 2\alpha x +\beta ) + 2\alpha v^2 \\\\\\\sf\longrightarrow a = \dfrac{ -\alpha v^2}{2\alpha x + \beta ) } \\\\\\\sf\longrightarrow a = -\alpha v^2 . v \\\\\\\sf\longrightarrow\boxed{ \red{\sf accl^n = -\alpha v^3 }}$

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