The distance $x$ covered by a particle in one-dimensional motion varies with time $t$ as $x^{2} = at^{2} + 2bt + c$. The accelerating of the particle varies as

Question

Dr. Abubacker · Accepted Answer

(a) $x^{2} = at^{2} + 2bt + c$
Differentiating w.r.t. time, we get
$2x\frac{dx}{dt} = 2at + 2b$ or $xv = at + b$ or $v = \frac{at + b}{x}$
Again differentiating w.r.t. time, we get
$x\ \frac{dx}{dt} + v\ \frac{dx}{dt} = a$
or $xA + v^{2} = a\ \ $or $xA = a - v^{2}$
or $xA = a - \left( \frac{at + b}{x} \right)^{2}$ or $xA = \frac{ax^{2} - a^{2}t^{2} - b^{2} - 2abt}{x^{2}}$
or $xA = \frac{a^{2}t^{2} + 2abt + ac - a^{2}t^{2} - b^{2} - 2abt}{x^{2}}$
or $A = \frac{ac - b^{2}}{x^{3}}$ or $A \propto x^{- 3}$

Q. The distance $x$ covered by a particle in one-dimensional motion varies with time $t$ as $x^{2} = at^{2} + 2bt + c$. The accelerating of the particle varies as

Explanation

Reviewed by Dr. Abubacker

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