Question

If the velocity of a body related to displacement $x$ is given by $v=\sqrt{5000+24x}\text{m/s},$ then the acceleration of the body is _____${\text{m/s}}^2.$

Answer 1

Given :-
$v=\sqrt{5000+24x}\text{m/s}$

Differentiating both side with respect to $x$,

$\frac{dv}{dx}=\frac{1}{2\sqrt{5000+24x}}\times 24$

$\frac{dv}{dx}=\frac{12}{\sqrt{5000+24x}}$

Multiplying both sides by $v$

$\frac{vdv}{dx}=\frac{12v}{\sqrt{5000+24x}}$

$a=\frac{12\sqrt{5000+24x}}{\sqrt{5000+24x}}(\because \frac{vdv}{dx}=a)$

$\therefore a=12{\text{m/s}}^2$

Correct answer $:12$