the speed v of particle moving along a straight line is given by $a$+b$y^{2} = x^{2}$(where $x$is its distance from the origin).The acceleration of the particle is

Question

Dr. Abubacker · Accepted Answer

(c)
Given equation is $a + bv^{2} = x^{2}$
On differentiating with respect to $t$, we get
$0 + b\left( 2v\frac{dv}{dt} \right) = 2x\frac{dx}{dt}$
$\Rightarrow \ \ \ vb\frac{dv}{dt} = x\frac{dx}{dt}$
$\Rightarrow \ \frac{dv}{dt} = \frac{x}{vb}.\frac{dx}{dt}$
$\Rightarrow \ \frac{dv}{dt} = \frac{x}{b}\left( \because\ \ \frac{dx}{dt} = v \right)$

Q. the speed v of particle moving along a straight line is given by $a$+b$y^{2} = x^{2}$(where $x$is its distance from the origin).The acceleration of the particle is

Explanation

Reviewed by Dr. Abubacker

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