First slide

Instantaneous acceleration

Question

The velocity v of a particle as a function of its position (x) is expressed as $v = \sqrt{c_{1} - c_{2} x}$ where c₁ and c₂ are positive constants. The acceleration of the particle is

Moderate

A

c₂
B

$- \frac{c_{2}}{2}$
C

$c_{1} - c_{2}$
D

$\frac{c_{1} + c_{2}}{2}$

Solution

Given, $v = \sqrt{c_{1} - c_{2} x}$
we know that, $a = v \frac{d v}{d x}$
$\Rightarrow$ $a = \sqrt{c_{1} - c_{2} x} \cdot \frac{d}{d x} 9 (\sqrt{c_{1} - c_{2} x})$
$= \frac{1}{2} (- c_{2}) = - \frac{c_{2}}{2}$

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